このページは、「シグマ記号の性質と公式」の練習問題アーカイブページとなります。
この問題の解き方の詳細は↓
シグマ記号の性質と公式 で確認できます。
問題アーカイブ01
問題アーカイブ01次の和を求めよ。
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}(4k+3)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n}(3k^2-7k+4)\)
\({\small (3)}~\) \(\displaystyle \sum_{k=1}^{n}k(k+2)\)
\({\small (4)}~\) \(\displaystyle \sum_{k=1}^{n-1}5k\)
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}(4k+3)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n}(3k^2-7k+4)\)
\({\small (3)}~\) \(\displaystyle \sum_{k=1}^{n}k(k+2)\)
\({\small (4)}~\) \(\displaystyle \sum_{k=1}^{n-1}5k\)
数研出版｜数学Ｂ[710] p.25 練習26
\({\small (1)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(4k+3)
\\[5pt]~~~&=&4\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}3
\\[5pt]~~~&=&4\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+3n
\\[5pt]~~~&=&2n(n+1)+3n
\\[5pt]~~~&=&2n^2+2n+3n
\\[5pt]~~~&=&2n^2+5n
\\[5pt]~~~&=&n(2n+5)
\end{eqnarray}\)
\({\small (2)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(3k^2-7k+4)
\\[5pt]~~~&=&3\displaystyle \sum_{k=1}^{n}k^2-7\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}4
\\[5pt]~~~&=&3\cdot \displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)-7\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+4n
\end{eqnarray}\)
\\[5pt]~~~&=&3\displaystyle \sum_{k=1}^{n}k^2-7\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}4
\\[5pt]~~~&=&3\cdot \displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)-7\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+4n
\end{eqnarray}\)
通分して、共通因数 \(\displaystyle \frac{\,1\,}{\,2\,}n\) でくくると、
\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(n+1)(2n+1)-\displaystyle \frac{\,7\,}{\,2\,}n(n+1)+\displaystyle \frac{\,8\,}{\,2\,}n
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left\{(n+1)(2n+1)-7(n+1)+8\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left(2n^2+3n+1-7n-7+8\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left(2n^2-4n+2\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\cdot 2\left(n^2-2n+1\right)
\\[5pt]~~~&=&n(n-1)^2
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left\{(n+1)(2n+1)-7(n+1)+8\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left(2n^2+3n+1-7n-7+8\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left(2n^2-4n+2\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\cdot 2\left(n^2-2n+1\right)
\\[5pt]~~~&=&n(n-1)^2
\end{eqnarray}\)
\({\small (3)}~\)展開してから、シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}k(k+2)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}(k^2+2k)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^2+2\displaystyle \sum_{k=1}^{n}k
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+2\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}(k^2+2k)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^2+2\displaystyle \sum_{k=1}^{n}k
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+2\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)
\end{eqnarray}\)
共通因数 \(\displaystyle \frac{\,1\,}{\,6\,}n(n+1)\) でくくると、
\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+\displaystyle \frac{\,6\,}{\,6\,}n(n+1)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)\left(2n+1+6\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+7)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)\left(2n+1+6\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+7)
\end{eqnarray}\)
\({\small (4)}~\)\(\begin{eqnarray}~~~\displaystyle \sum_{k=1}^{n-1}5k&=&5\displaystyle \sum_{k=1}^{n-1}k\end{eqnarray}\)
公式 \(\displaystyle \sum_{k=1}^{n} k=\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\) を \(n \to n-1\) に書き換えると、
\(\begin{eqnarray}\hspace{25pt}~~~&=&5\cdot \displaystyle \frac{\,1\,}{\,2\,}(n-1)\left\{(n-1)+1\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,5\,}{\,2\,}(n-1)\cdot n
\\[5pt]~~~&=&\displaystyle \frac{\,5\,}{\,2\,}n(n-1)
\end{eqnarray}\)
問題アーカイブ02
問題アーカイブ02次の和を求めよ。
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}(4k-5)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n}(3k^2-7k+4)\)
\({\small (3)}~\) \(\displaystyle \sum_{k=1}^{n}(k^3+k)\)
\({\small (4)}~\) \(\displaystyle \sum_{k=1}^{n-1}2k\)
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}(4k-5)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n}(3k^2-7k+4)\)
\({\small (3)}~\) \(\displaystyle \sum_{k=1}^{n}(k^3+k)\)
\({\small (4)}~\) \(\displaystyle \sum_{k=1}^{n-1}2k\)
数研出版｜高等学校数学Ｂ[711] p.28 練習28
\({\small (1)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(4k-5)
\\[5pt]~~~&=&4\displaystyle \sum_{k=1}^{n}k-\displaystyle \sum_{k=1}^{n}5
\\[5pt]~~~&=&4\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)-5n
\\[5pt]~~~&=&2n(n+1)-5n
\\[5pt]~~~&=&2n^2+2n-5n
\\[5pt]~~~&=&2n^2-3n
\\[5pt]~~~&=&n(2n-3)
\end{eqnarray}\)
\({\small (2)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(3k^2-7k+4)
\\[5pt]~~~&=&3\displaystyle \sum_{k=1}^{n}k^2-7\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}4
\\[5pt]~~~&=&3\cdot \displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)-7\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+4n
\end{eqnarray}\)
\\[5pt]~~~&=&3\displaystyle \sum_{k=1}^{n}k^2-7\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}4
\\[5pt]~~~&=&3\cdot \displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)-7\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+4n
\end{eqnarray}\)
通分して、共通因数 \(\displaystyle \frac{\,1\,}{\,2\,}n\) でくくると、
\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(n+1)(2n+1)-\displaystyle \frac{\,7\,}{\,2\,}n(n+1)+\displaystyle \frac{\,8\,}{\,2\,}n
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left\{(n+1)(2n+1)-7(n+1)+8\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left(2n^2+3n+1-7n-7+8\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left(2n^2-4n+2\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\cdot 2\left(n^2-2n+1\right)
\\[5pt]~~~&=&n(n-1)^2
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left\{(n+1)(2n+1)-7(n+1)+8\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left(2n^2+3n+1-7n-7+8\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left(2n^2-4n+2\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\cdot 2\left(n^2-2n+1\right)
\\[5pt]~~~&=&n(n-1)^2
\end{eqnarray}\)
\({\small (3)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(k^3+k)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^3+\displaystyle \sum_{k=1}^{n}k
\\[5pt]~~~&=&\left\{\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\right\}^2+\displaystyle \frac{\,1\,}{\,2\,}n(n+1)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^3+\displaystyle \sum_{k=1}^{n}k
\\[5pt]~~~&=&\left\{\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\right\}^2+\displaystyle \frac{\,1\,}{\,2\,}n(n+1)
\end{eqnarray}\)
共通因数 \(\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\) でくくると、
\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\left\{\displaystyle \frac{\,1\,}{\,2\,}n(n+1)+1\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\cdot \displaystyle \frac{\,1\,}{\,2\,}\left\{n(n+1)+2\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)\left(n^2+n+2\right)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\cdot \displaystyle \frac{\,1\,}{\,2\,}\left\{n(n+1)+2\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)\left(n^2+n+2\right)
\end{eqnarray}\)
\({\small (4)}~\)\(\begin{eqnarray}~~~\displaystyle \sum_{k=1}^{n-1}2k&=&2\displaystyle \sum_{k=1}^{n-1}k\end{eqnarray}\)
公式 \(\displaystyle \sum_{k=1}^{n} k=\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\) を \(n \to n-1\) に書き換えると、
\(\begin{eqnarray}\hspace{25pt}~~~&=&2\cdot \displaystyle \frac{\,1\,}{\,2\,}(n-1)\left\{(n-1)+1\right\}
\\[5pt]~~~&=&(n-1)\cdot n
\\[5pt]~~~&=&n(n-1)
\end{eqnarray}\)
問題アーカイブ03
問題アーカイブ03次の和を求めよ。
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}(3^k+2k+1)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n}(k-1)(k+2)\)
\({\small (3)}~\) \(\displaystyle \sum_{k=1}^{n}(k^3-k)\)
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}(3^k+2k+1)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n}(k-1)(k+2)\)
\({\small (3)}~\) \(\displaystyle \sum_{k=1}^{n}(k^3-k)\)
数研出版｜高等学校数学Ｂ[711] p.34 問題 9
\({\small (1)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(3^k+2k+1)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}3^k+2\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}1
\\[5pt]~~~&=&\displaystyle \frac{\,3(3^n-1)\,}{\,3-1\,}+2\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+n
\\[5pt]~~~&=&\displaystyle \frac{\,3(3^n-1)\,}{\,2\,}+n(n+1)+n
\\[5pt]~~~&=&\displaystyle \frac{\,3^{n+1}-3\,}{\,2\,}+n^2+2n
\\[5pt]~~~&=&\displaystyle \frac{\,3^{n+1}-3+2n^2+4n\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,3^{n+1}+2n^2+4n-3\,}{\,2\,}
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}3^k+2\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}1
\\[5pt]~~~&=&\displaystyle \frac{\,3(3^n-1)\,}{\,3-1\,}+2\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+n
\\[5pt]~~~&=&\displaystyle \frac{\,3(3^n-1)\,}{\,2\,}+n(n+1)+n
\\[5pt]~~~&=&\displaystyle \frac{\,3^{n+1}-3\,}{\,2\,}+n^2+2n
\\[5pt]~~~&=&\displaystyle \frac{\,3^{n+1}-3+2n^2+4n\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,3^{n+1}+2n^2+4n-3\,}{\,2\,}
\end{eqnarray}\)
\({\small (2)}~\)展開してから、シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(k-1)(k+2)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}(k^2+k-2)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^2+\displaystyle \sum_{k=1}^{n}k-\displaystyle \sum_{k=1}^{n}2
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+\displaystyle \frac{\,1\,}{\,2\,}n(n+1)-2n
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}(k^2+k-2)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^2+\displaystyle \sum_{k=1}^{n}k-\displaystyle \sum_{k=1}^{n}2
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+\displaystyle \frac{\,1\,}{\,2\,}n(n+1)-2n
\end{eqnarray}\)
通分して、共通因数 \(\displaystyle \frac{\,1\,}{\,6\,}n\) でくくると、
\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+\displaystyle \frac{\,3\,}{\,6\,}n(n+1)-\displaystyle \frac{\,12\,}{\,6\,}n
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left\{(n+1)(2n+1)+3(n+1)-12\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(2n^2+3n+1+3n+3-12\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(2n^2+6n-8\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\cdot 2\left(n^2+3n-4\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,3\,}n(n+4)(n-1)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left\{(n+1)(2n+1)+3(n+1)-12\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(2n^2+3n+1+3n+3-12\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(2n^2+6n-8\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\cdot 2\left(n^2+3n-4\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,3\,}n(n+4)(n-1)
\end{eqnarray}\)
\({\small (3)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(k^3-k)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^3-\displaystyle \sum_{k=1}^{n}k
\\[5pt]~~~&=&\left\{\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\right\}^2-\displaystyle \frac{\,1\,}{\,2\,}n(n+1)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^3-\displaystyle \sum_{k=1}^{n}k
\\[5pt]~~~&=&\left\{\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\right\}^2-\displaystyle \frac{\,1\,}{\,2\,}n(n+1)
\end{eqnarray}\)
共通因数 \(\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\) でくくると、
\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\left\{\displaystyle \frac{\,1\,}{\,2\,}n(n+1)-1\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\cdot \displaystyle \frac{\,1\,}{\,2\,}\left\{n(n+1)-2\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)\left(n^2+n-2\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)(n+2)(n-1)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}(n-1)n(n+1)(n+2)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\cdot \displaystyle \frac{\,1\,}{\,2\,}\left\{n(n+1)-2\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)\left(n^2+n-2\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)(n+2)(n-1)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}(n-1)n(n+1)(n+2)
\end{eqnarray}\)
問題アーカイブ04
問題アーカイブ04次の和を求めよ。
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}(2k+1)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n}(3k-5)\)
\({\small (3)}~\) \(\displaystyle \sum_{k=1}^{n-1}4k\)
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}(2k+1)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n}(3k-5)\)
\({\small (3)}~\) \(\displaystyle \sum_{k=1}^{n-1}4k\)
数研出版｜新編数学Ｂ[712] p.26 練習28
\({\small (1)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(2k+1)
\\[5pt]~~~&=&2\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}1
\\[5pt]~~~&=&2\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+n
\\[5pt]~~~&=&n(n+1)+n
\\[5pt]~~~&=&n^2+n+n
\\[5pt]~~~&=&n^2+2n
\\[5pt]~~~&=&n(n+2)
\end{eqnarray}\)
\({\small (2)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(3k-5)
\\[5pt]~~~&=&3\displaystyle \sum_{k=1}^{n}k-\displaystyle \sum_{k=1}^{n}5
\\[5pt]~~~&=&3\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)-5n
\\[5pt]~~~&=&\displaystyle \frac{\,3\,}{\,2\,}n(n+1)-5n
\\[5pt]~~~&=&\displaystyle \frac{\,3\,}{\,2\,}n(n+1)-\displaystyle \frac{\,10\,}{\,2\,}n
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left\{3(n+1)-10\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(3n+3-10)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(3n-7)
\end{eqnarray}\)
\({\small (3)}~\)\(\begin{eqnarray}~~~\displaystyle \sum_{k=1}^{n-1}4k&=&4\displaystyle \sum_{k=1}^{n-1}k\end{eqnarray}\)
公式 \(\displaystyle \sum_{k=1}^{n} k=\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\) を \(n \to n-1\) に書き換えると、
\(\begin{eqnarray}\hspace{25pt}~~~&=&4\cdot \displaystyle \frac{\,1\,}{\,2\,}(n-1)\left\{(n-1)+1\right\}
\\[5pt]~~~&=&2(n-1)\cdot n
\\[5pt]~~~&=&2n(n-1)
\end{eqnarray}\)
問題アーカイブ05
問題アーカイブ05次の和を求めよ。
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}(3k^2-7k+4)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n}(k-1)(k-2)\)
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}(3k^2-7k+4)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n}(k-1)(k-2)\)
数研出版｜新編数学Ｂ[712] p.27 練習29
\({\small (1)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(3k^2-7k+4)
\\[5pt]~~~&=&3\displaystyle \sum_{k=1}^{n}k^2-7\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}4
\\[5pt]~~~&=&3\cdot \displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)-7\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+4n
\end{eqnarray}\)
\\[5pt]~~~&=&3\displaystyle \sum_{k=1}^{n}k^2-7\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}4
\\[5pt]~~~&=&3\cdot \displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)-7\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+4n
\end{eqnarray}\)
通分して、共通因数 \(\displaystyle \frac{\,1\,}{\,2\,}n\) でくくると、
\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(n+1)(2n+1)-\displaystyle \frac{\,7\,}{\,2\,}n(n+1)+\displaystyle \frac{\,8\,}{\,2\,}n
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left\{(n+1)(2n+1)-7(n+1)+8\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left(2n^2+3n+1-7n-7+8\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left(2n^2-4n+2\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\cdot 2\left(n^2-2n+1\right)
\\[5pt]~~~&=&n(n-1)^2
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left\{(n+1)(2n+1)-7(n+1)+8\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left(2n^2+3n+1-7n-7+8\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left(2n^2-4n+2\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\cdot 2\left(n^2-2n+1\right)
\\[5pt]~~~&=&n(n-1)^2
\end{eqnarray}\)
\({\small (2)}~\)展開してから、シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(k-1)(k-2)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}(k^2-3k+2)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^2-3\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}2
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)-3\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+2n
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}(k^2-3k+2)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^2-3\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}2
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)-3\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+2n
\end{eqnarray}\)
通分して、共通因数 \(\displaystyle \frac{\,1\,}{\,6\,}n\) でくくると、
\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)-\displaystyle \frac{\,9\,}{\,6\,}n(n+1)+\displaystyle \frac{\,12\,}{\,6\,}n
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left\{(n+1)(2n+1)-9(n+1)+12\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(2n^2+3n+1-9n-9+12\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(2n^2-6n+4\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\cdot 2\left(n^2-3n+2\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,3\,}n(n-1)(n-2)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left\{(n+1)(2n+1)-9(n+1)+12\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(2n^2+3n+1-9n-9+12\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(2n^2-6n+4\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\cdot 2\left(n^2-3n+2\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,3\,}n(n-1)(n-2)
\end{eqnarray}\)
問題アーカイブ06
問題アーカイブ06和 \(\displaystyle \sum_{k=1}^{n}k(k+1)(k+2)\) を求めよ。ただし、問題5の公式を用いてよい。
数研出版｜新編数学Ｂ[712] p.34 補充問題 6
展開してから、シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}k(k+1)(k+2)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k(k^2+3k+2)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}(k^3+3k^2+2k)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^3+3\displaystyle \sum_{k=1}^{n}k^2+2\displaystyle \sum_{k=1}^{n}k
\\[5pt]~~~&=&\left\{\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\right\}^2+3\cdot \displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+2\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n^2(n+1)^2+\displaystyle \frac{\,1\,}{\,2\,}n(n+1)(2n+1)+n(n+1)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k(k^2+3k+2)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}(k^3+3k^2+2k)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^3+3\displaystyle \sum_{k=1}^{n}k^2+2\displaystyle \sum_{k=1}^{n}k
\\[5pt]~~~&=&\left\{\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\right\}^2+3\cdot \displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+2\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n^2(n+1)^2+\displaystyle \frac{\,1\,}{\,2\,}n(n+1)(2n+1)+n(n+1)
\end{eqnarray}\)
通分して、共通因数 \(\displaystyle \frac{\,1\,}{\,4\,}n(n+1)\) でくくると、
\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n^2(n+1)^2+\displaystyle \frac{\,2\,}{\,4\,}n(n+1)(2n+1)+\displaystyle \frac{\,4\,}{\,4\,}n(n+1)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)\left\{n(n+1)+2(2n+1)+4\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)\left(n^2+n+4n+2+4\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)\left(n^2+5n+6\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)(n+2)(n+3)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)\left\{n(n+1)+2(2n+1)+4\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)\left(n^2+n+4n+2+4\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)\left(n^2+5n+6\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)(n+2)(n+3)
\end{eqnarray}\)
問題アーカイブ07
問題アーカイブ07次の和を求めよ。
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}(5k+1)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n}(k+1)(k-2)\)
\({\small (3)}~\) \(\displaystyle \sum_{k=1}^{n}(k^3-k)\)
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}(5k+1)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n}(k+1)(k-2)\)
\({\small (3)}~\) \(\displaystyle \sum_{k=1}^{n}(k^3-k)\)
東京書籍｜Advanced数学Ｂ[701] p.22 問29
\({\small (1)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(5k+1)
\\[5pt]~~~&=&5\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}1
\\[5pt]~~~&=&5\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+n
\\[5pt]~~~&=&\displaystyle \frac{\,5\,}{\,2\,}n(n+1)+n
\\[5pt]~~~&=&\displaystyle \frac{\,5\,}{\,2\,}n(n+1)+\displaystyle \frac{\,2\,}{\,2\,}n
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left\{5(n+1)+2\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(5n+5+2)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(5n+7)
\end{eqnarray}\)
\({\small (2)}~\)展開してから、シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(k+1)(k-2)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}(k^2-k-2)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^2-\displaystyle \sum_{k=1}^{n}k-\displaystyle \sum_{k=1}^{n}2
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)-\displaystyle \frac{\,1\,}{\,2\,}n(n+1)-2n
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}(k^2-k-2)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^2-\displaystyle \sum_{k=1}^{n}k-\displaystyle \sum_{k=1}^{n}2
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)-\displaystyle \frac{\,1\,}{\,2\,}n(n+1)-2n
\end{eqnarray}\)
通分して、共通因数 \(\displaystyle \frac{\,1\,}{\,6\,}n\) でくくると、
\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)-\displaystyle \frac{\,3\,}{\,6\,}n(n+1)-\displaystyle \frac{\,12\,}{\,6\,}n
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left\{(n+1)(2n+1)-3(n+1)-12\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(2n^2+3n+1-3n-3-12\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(2n^2-14\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\cdot 2\left(n^2-7\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,3\,}n(n^2-7)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left\{(n+1)(2n+1)-3(n+1)-12\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(2n^2+3n+1-3n-3-12\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(2n^2-14\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\cdot 2\left(n^2-7\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,3\,}n(n^2-7)
\end{eqnarray}\)
\({\small (3)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(k^3-k)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^3-\displaystyle \sum_{k=1}^{n}k
\\[5pt]~~~&=&\left\{\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\right\}^2-\displaystyle \frac{\,1\,}{\,2\,}n(n+1)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^3-\displaystyle \sum_{k=1}^{n}k
\\[5pt]~~~&=&\left\{\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\right\}^2-\displaystyle \frac{\,1\,}{\,2\,}n(n+1)
\end{eqnarray}\)
共通因数 \(\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\) でくくると、
\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\left\{\displaystyle \frac{\,1\,}{\,2\,}n(n+1)-1\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\cdot \displaystyle \frac{\,1\,}{\,2\,}\left\{n(n+1)-2\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)\left(n^2+n-2\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)(n+2)(n-1)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}(n-1)n(n+1)(n+2)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\cdot \displaystyle \frac{\,1\,}{\,2\,}\left\{n(n+1)-2\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)\left(n^2+n-2\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)(n+2)(n-1)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}(n-1)n(n+1)(n+2)
\end{eqnarray}\)
問題アーカイブ08
問題アーカイブ08次の和を求めよ。
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}(3k+2)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n-1}(2k-1)\)
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}(3k+2)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n-1}(2k-1)\)
東京書籍｜Standard数学Ｂ[702] p.31 問7
\({\small (1)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(3k+2)
\\[5pt]~~~&=&3\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}2
\\[5pt]~~~&=&3\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+2n
\\[5pt]~~~&=&\displaystyle \frac{\,3\,}{\,2\,}n(n+1)+2n
\\[5pt]~~~&=&\displaystyle \frac{\,3\,}{\,2\,}n(n+1)+\displaystyle \frac{\,4\,}{\,2\,}n
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n\left\{3(n+1)+4\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(3n+3+4)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(3n+7)
\end{eqnarray}\)
\({\small (2)}~\)
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n-1}(2k-1)
\\[5pt]~~~&=&2\displaystyle \sum_{k=1}^{n-1}k-\displaystyle \sum_{k=1}^{n-1}1\end{eqnarray}\)
公式 \(\displaystyle \sum_{k=1}^{n} k=\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\) と \(\displaystyle \sum_{k=1}^{n} 1=n\) を \(n \to n-1\) に書き換えると、
\(\begin{eqnarray}\hspace{25pt}~~~&=&2\cdot \displaystyle \frac{\,1\,}{\,2\,}(n-1)\left\{(n-1)+1\right\}-(n-1)
\\[5pt]~~~&=&(n-1)\cdot n-(n-1)
\\[5pt]~~~&=&(n-1)(n-1)
\\[5pt]~~~&=&(n-1)^2
\end{eqnarray}\)
問題アーカイブ09
問題アーカイブ09次の和を求めよ。
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}2k(3k+2)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n-1}(3k+2)(k-1)\)
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}2k(3k+2)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n-1}(3k+2)(k-1)\)
東京書籍｜Standard数学Ｂ[702] p.32 問8
\({\small (1)}~\)展開してから、シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}2k(3k+2)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}(6k^2+4k)
\\[5pt]~~~&=&6\displaystyle \sum_{k=1}^{n}k^2+4\displaystyle \sum_{k=1}^{n}k
\\[5pt]~~~&=&6\cdot \displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+4\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)
\\[5pt]~~~&=&n(n+1)(2n+1)+2n(n+1)
\\[5pt]~~~&=&n(n+1)\left\{(2n+1)+2\right\}
\\[5pt]~~~&=&n(n+1)(2n+3)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}(6k^2+4k)
\\[5pt]~~~&=&6\displaystyle \sum_{k=1}^{n}k^2+4\displaystyle \sum_{k=1}^{n}k
\\[5pt]~~~&=&6\cdot \displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+4\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)
\\[5pt]~~~&=&n(n+1)(2n+1)+2n(n+1)
\\[5pt]~~~&=&n(n+1)\left\{(2n+1)+2\right\}
\\[5pt]~~~&=&n(n+1)(2n+3)
\end{eqnarray}\)
\({\small (2)}~\)展開してから、シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n-1}(3k+2)(k-1)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n-1}(3k^2-k-2)
\\[5pt]~~~&=&3\displaystyle \sum_{k=1}^{n-1}k^2-\displaystyle \sum_{k=1}^{n-1}k-\displaystyle \sum_{k=1}^{n-1}2
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n-1}(3k^2-k-2)
\\[5pt]~~~&=&3\displaystyle \sum_{k=1}^{n-1}k^2-\displaystyle \sum_{k=1}^{n-1}k-\displaystyle \sum_{k=1}^{n-1}2
\end{eqnarray}\)
各公式を \(n \to n-1\) に書き換えると、
\(\begin{eqnarray}~~~&=&3\cdot \displaystyle \frac{\,1\,}{\,6\,}(n-1)n(2n-1)-\displaystyle \frac{\,1\,}{\,2\,}(n-1)n-2(n-1)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(n-1)n(2n-1)-\displaystyle \frac{\,1\,}{\,2\,}(n-1)n-2(n-1)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(n-1)n(2n-1)-\displaystyle \frac{\,1\,}{\,2\,}(n-1)n-2(n-1)
\end{eqnarray}\)
通分して、共通因数 \(\displaystyle \frac{\,1\,}{\,2\,}(n-1)\) でくくると、
\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(n-1)n(2n-1)-\displaystyle \frac{\,1\,}{\,2\,}(n-1)n-\displaystyle \frac{\,4\,}{\,2\,}(n-1)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(n-1)\left\{n(2n-1)-n-4\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(n-1)\left(2n^2-n-n-4\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(n-1)\left(2n^2-2n-4\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(n-1)\cdot 2\left(n^2-n-2\right)
\\[5pt]~~~&=&(n-1)(n+1)(n-2)
\\[5pt]~~~&=&(n-2)(n-1)(n+1)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(n-1)\left\{n(2n-1)-n-4\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(n-1)\left(2n^2-n-n-4\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(n-1)\left(2n^2-2n-4\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(n-1)\cdot 2\left(n^2-n-2\right)
\\[5pt]~~~&=&(n-1)(n+1)(n-2)
\\[5pt]~~~&=&(n-2)(n-1)(n+1)
\end{eqnarray}\)
問題アーカイブ10
問題アーカイブ10次の和を求めよ。
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}(4k+3)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n+1}(3k-5)\)
\({\small (3)}~\) \(\displaystyle \sum_{j=1}^{n}(-12j+6)\)
\({\small (4)}~\) \(\displaystyle \sum_{k=1}^{n}(k^2+2k+3)\)
\({\small (5)}~\) \(\displaystyle \sum_{k=1}^{n}(k+3)(2k-1)\)
\({\small (6)}~\) \(\displaystyle \sum_{k=1}^{n}4\cdot (-3)^{k-1}\)
\({\small (7)}~\) \(\displaystyle \sum_{k=1}^{n}7^k\)
\({\small (8)}~\) \(\displaystyle \sum_{k=1}^{n-1}(2^k-k^2)\)
\({\small (9)}~\) \(\displaystyle \sum_{k=1}^{n}(k^3-k)\)
\({\small (1)}~\) \(\displaystyle \sum_{k=1}^{n}(4k+3)\)
\({\small (2)}~\) \(\displaystyle \sum_{k=1}^{n+1}(3k-5)\)
\({\small (3)}~\) \(\displaystyle \sum_{j=1}^{n}(-12j+6)\)
\({\small (4)}~\) \(\displaystyle \sum_{k=1}^{n}(k^2+2k+3)\)
\({\small (5)}~\) \(\displaystyle \sum_{k=1}^{n}(k+3)(2k-1)\)
\({\small (6)}~\) \(\displaystyle \sum_{k=1}^{n}4\cdot (-3)^{k-1}\)
\({\small (7)}~\) \(\displaystyle \sum_{k=1}^{n}7^k\)
\({\small (8)}~\) \(\displaystyle \sum_{k=1}^{n-1}(2^k-k^2)\)
\({\small (9)}~\) \(\displaystyle \sum_{k=1}^{n}(k^3-k)\)
東京書籍｜Standard数学Ｂ[702] p.40 Training 11
\({\small (1)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(4k+3)
\\[5pt]~~~&=&4\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}3
\\[5pt]~~~&=&4\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+3n
\\[5pt]~~~&=&2n(n+1)+3n
\\[5pt]~~~&=&2n^2+2n+3n
\\[5pt]~~~&=&2n^2+5n
\\[5pt]~~~&=&n(2n+5)
\end{eqnarray}\)
\({\small (2)}~\)
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n+1}(3k-5)
\\[5pt]~~~&=&3\displaystyle \sum_{k=1}^{n+1}k-\displaystyle \sum_{k=1}^{n+1}5\end{eqnarray}\)
各公式を \(n \to n+1\) に書き換えると、
\(\begin{eqnarray}~~~&=&3\cdot \displaystyle \frac{\,1\,}{\,2\,}(n+1)\left\{(n+1)+1\right\}-5(n+1)
\\[5pt]~~~&=&\displaystyle \frac{\,3\,}{\,2\,}(n+1)(n+2)-5(n+1)
\\[5pt]~~~&=&\displaystyle \frac{\,3\,}{\,2\,}(n+1)(n+2)-\displaystyle \frac{\,10\,}{\,2\,}(n+1)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(n+1)\left\{3(n+2)-10\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(n+1)(3n+6-10)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}(n+1)(3n-4)
\end{eqnarray}\)
\({\small (3)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{j=1}^{n}(-12j+6)
\\[5pt]~~~&=&-12\displaystyle \sum_{j=1}^{n}j+\displaystyle \sum_{j=1}^{n}6
\\[5pt]~~~&=&-12\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+6n
\\[5pt]~~~&=&-6n(n+1)+6n
\\[5pt]~~~&=&6n\left\{-(n+1)+1\right\}
\\[5pt]~~~&=&6n(-n)
\\[5pt]~~~&=&-6n^2
\end{eqnarray}\)
\({\small (4)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(k^2+2k+3)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^2+2\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}3
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+2\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+3n
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+n(n+1)+3n
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^2+2\displaystyle \sum_{k=1}^{n}k+\displaystyle \sum_{k=1}^{n}3
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+2\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)+3n
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+n(n+1)+3n
\end{eqnarray}\)
通分して、共通因数 \(\displaystyle \frac{\,1\,}{\,6\,}n\) でくくると、
\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+\displaystyle \frac{\,6\,}{\,6\,}n(n+1)+\displaystyle \frac{\,18\,}{\,6\,}n
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left\{(n+1)(2n+1)+6(n+1)+18\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(2n^2+3n+1+6n+6+18\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(2n^2+9n+25\right)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left\{(n+1)(2n+1)+6(n+1)+18\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(2n^2+3n+1+6n+6+18\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(2n^2+9n+25\right)
\end{eqnarray}\)
\({\small (5)}~\)展開してから、シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(k+3)(2k-1)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}(2k^2+5k-3)
\\[5pt]~~~&=&2\displaystyle \sum_{k=1}^{n}k^2+5\displaystyle \sum_{k=1}^{n}k-\displaystyle \sum_{k=1}^{n}3
\\[5pt]~~~&=&2\cdot \displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+5\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)-3n
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,3\,}n(n+1)(2n+1)+\displaystyle \frac{\,5\,}{\,2\,}n(n+1)-3n
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}(2k^2+5k-3)
\\[5pt]~~~&=&2\displaystyle \sum_{k=1}^{n}k^2+5\displaystyle \sum_{k=1}^{n}k-\displaystyle \sum_{k=1}^{n}3
\\[5pt]~~~&=&2\cdot \displaystyle \frac{\,1\,}{\,6\,}n(n+1)(2n+1)+5\cdot \displaystyle \frac{\,1\,}{\,2\,}n(n+1)-3n
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,3\,}n(n+1)(2n+1)+\displaystyle \frac{\,5\,}{\,2\,}n(n+1)-3n
\end{eqnarray}\)
通分して、共通因数 \(\displaystyle \frac{\,1\,}{\,6\,}n\) でくくると、
\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,2\,}{\,6\,}n(n+1)(2n+1)+\displaystyle \frac{\,15\,}{\,6\,}n(n+1)-\displaystyle \frac{\,18\,}{\,6\,}n
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left\{2(n+1)(2n+1)+15(n+1)-18\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(4n^2+6n+2+15n+15-18\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(4n^2+21n-1\right)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left\{2(n+1)(2n+1)+15(n+1)-18\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(4n^2+6n+2+15n+15-18\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,6\,}n\left(4n^2+21n-1\right)
\end{eqnarray}\)
\({\small (6)}~\)等比数列の和の公式を用いると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}4\cdot (-3)^{k-1}
\\[5pt]~~~&=&4\displaystyle \sum_{k=1}^{n}(-3)^{k-1}
\\[5pt]~~~&=&4\cdot \displaystyle \frac{\,1\cdot \left\{(-3)^n-1\right\}\,}{\,-3-1\,}
\\[5pt]~~~&=&4\cdot \displaystyle \frac{\,(-3)^n-1\,}{\,-4\,}
\\[5pt]~~~&=&-\left\{(-3)^n-1\right\}
\\[5pt]~~~&=&1-(-3)^n
\end{eqnarray}\)
\\[5pt]~~~&=&4\displaystyle \sum_{k=1}^{n}(-3)^{k-1}
\\[5pt]~~~&=&4\cdot \displaystyle \frac{\,1\cdot \left\{(-3)^n-1\right\}\,}{\,-3-1\,}
\\[5pt]~~~&=&4\cdot \displaystyle \frac{\,(-3)^n-1\,}{\,-4\,}
\\[5pt]~~~&=&-\left\{(-3)^n-1\right\}
\\[5pt]~~~&=&1-(-3)^n
\end{eqnarray}\)
\({\small (7)}~\)等比数列の和の公式を用いると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}7^k
\\[5pt]~~~&=&\displaystyle \frac{\,7(7^n-1)\,}{\,7-1\,}
\\[5pt]~~~&=&\displaystyle \frac{\,7(7^n-1)\,}{\,6\,}
\\[5pt]~~~&=&\displaystyle \frac{\,7^{n+1}-7\,}{\,6\,}
\end{eqnarray}\)
\({\small (8)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n-1}(2^k-k^2)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n-1}2^k-\displaystyle \sum_{k=1}^{n-1}k^2
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n-1}2^k-\displaystyle \sum_{k=1}^{n-1}k^2
\end{eqnarray}\)
各公式を \(n \to n-1\) に書き換えると、
\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,2(2^{n-1}-1)\,}{\,2-1\,}-\displaystyle \frac{\,1\,}{\,6\,}(n-1)n(2n-1)
\\[5pt]~~~&=&2^n-2-\displaystyle \frac{\,1\,}{\,6\,}(n-1)n(2n-1)
\\[5pt]~~~&=&\displaystyle \frac{\,6\cdot 2^n-12-(n-1)n(2n-1)\,}{\,6\,}
\\[5pt]~~~&=&\displaystyle \frac{\,6\cdot 2^n-12-(2n^3-3n^2+n)\,}{\,6\,}
\\[5pt]~~~&=&\displaystyle \frac{\,6\cdot 2^n-2n^3+3n^2-n-12\,}{\,6\,}
\\[5pt]~~~&=&\displaystyle \frac{\,-2n^3+3n^2-n+6\cdot 2^n-12\,}{\,6\,}
\end{eqnarray}\)
\\[5pt]~~~&=&2^n-2-\displaystyle \frac{\,1\,}{\,6\,}(n-1)n(2n-1)
\\[5pt]~~~&=&\displaystyle \frac{\,6\cdot 2^n-12-(n-1)n(2n-1)\,}{\,6\,}
\\[5pt]~~~&=&\displaystyle \frac{\,6\cdot 2^n-12-(2n^3-3n^2+n)\,}{\,6\,}
\\[5pt]~~~&=&\displaystyle \frac{\,6\cdot 2^n-2n^3+3n^2-n-12\,}{\,6\,}
\\[5pt]~~~&=&\displaystyle \frac{\,-2n^3+3n^2-n+6\cdot 2^n-12\,}{\,6\,}
\end{eqnarray}\)
\({\small (9)}~\)シグマ記号を分けて、それぞれ公式で計算すると、
\(\begin{eqnarray}~~~&&\displaystyle \sum_{k=1}^{n}(k^3-k)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^3-\displaystyle \sum_{k=1}^{n}k
\\[5pt]~~~&=&\left\{\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\right\}^2-\displaystyle \frac{\,1\,}{\,2\,}n(n+1)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \sum_{k=1}^{n}k^3-\displaystyle \sum_{k=1}^{n}k
\\[5pt]~~~&=&\left\{\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\right\}^2-\displaystyle \frac{\,1\,}{\,2\,}n(n+1)
\end{eqnarray}\)
共通因数 \(\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\) でくくると、
\(\begin{eqnarray}~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\left\{\displaystyle \frac{\,1\,}{\,2\,}n(n+1)-1\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\cdot \displaystyle \frac{\,1\,}{\,2\,}\left\{n(n+1)-2\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)\left(n^2+n-2\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)(n+2)(n-1)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}(n-1)n(n+1)(n+2)
\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}n(n+1)\cdot \displaystyle \frac{\,1\,}{\,2\,}\left\{n(n+1)-2\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)\left(n^2+n-2\right)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}n(n+1)(n+2)(n-1)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,4\,}(n-1)n(n+1)(n+2)
\end{eqnarray}\)
