部分分数に分ける分数数列の和

数研出版｜高等学校数学Ｂ[711] p.34 問題 12(2)
数研出版｜新編数学Ｂ[712] p.46 章末問題A 7(2)

\(\displaystyle \frac{2}{k(k+2)}\) を部分分数に分けた式は、

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\(\begin{eqnarray}~~~\frac{1}{k}-\frac{1}{k+2}&=&\frac{(k+2)-k}{k(k+2)}
\\[5pt]~~~&=&\frac{2}{k(k+2)}
\end{eqnarray}\)

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よって、恒等式

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\(\begin{eqnarray}~~~\displaystyle \frac{2}{k(k+2)}=\left(\displaystyle \frac{1}{k}-\displaystyle \frac{1}{k+2}\right)\end{eqnarray}\)

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が成り立つ

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これを用いて、

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\(\begin{eqnarray}~~~&&\sum_{k=1}^{n}\frac{2}{k(k+2)}
\\[5pt]~~~&=&\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+2}\right)
\end{eqnarray}\)

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ここで、\(k=1\sim n\) までの自然数を代入した和を書き並べると、

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\(\begin{eqnarray} \require{cancel}~~~=&&\left(\frac{\,1\,}{\,1\,}-\cancel{\frac{\,1\,}{\,3\,}}\right)
\\[5pt]~~~+&&\left(\frac{\,1\,}{\,2\,}-\cancel{\frac{\,1\,}{\,4\,}}\right)
\\[5pt]~~~+&&\left(\cancel{\frac{\,1\,}{\,3\,}}-\cancel{\frac{\,1\,}{\,5\,}}\right)
\\[5pt]~~~+&&\left(\cancel{\frac{\,1\,}{\,4\,}}-\cancel{\frac{\,1\,}{\,6\,}}\right)
\\[5pt]~~~&& ~ ~ \cdots
\\[5pt]~~~+&&\left(\cancel{\frac{\,1\,}{\,n-1\,}}-\frac{\,1\,}{\,n+1\,}\right)
\\[5pt]~~~+&&\left(\cancel{\frac{\,1\,}{\,n\,}}-\frac{\,1\,}{\,n+2\,}\right)
\end{eqnarray}\)

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消えていく項と残る項を確認すると、

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\(\begin{eqnarray}~~~&=&1+\frac{\,1\,}{\,2\,}-\frac{\,1\,}{\,n+1\,}-\frac{\,1\,}{\,n+2\,}
\\[5pt]~~~&=&\frac{\,3\,}{\,2\,}-\frac{\,1\,}{\,n+1\,}-\frac{\,1\,}{\,n+2\,}
\\[5pt]~~~&=&\frac{\,3\,}{\,2\,}-\frac{\,(n+2)+(n+1)\,}{\,(n+1)(n+2)\,}
\\[5pt]~~~&=&\frac{\,3\,}{\,2\,}-\frac{\,2n+3\,}{\,(n+1)(n+2)\,}
\\[5pt]~~~&=&\frac{\,3(n+1)(n+2)-2(2n+3)\,}{\,2(n+1)(n+2)\,}
\\[5pt]~~~&=&\frac{\,3n^2+9n+6-4n-6\,}{\,2(n+1)(n+2)\,}
\\[5pt]~~~&=&\frac{\,3n^2+5n\,}{\,2(n+1)(n+2)\,}
\\[5pt]~~~&=&\frac{\,n(3n+5)\,}{\,2(n+1)(n+2)\,}
\end{eqnarray}\)

数研出版｜新編数学Ｂ[712] p.34 補充問題 7(1)

\(\displaystyle \frac{1}{k^2+3k+2}\) の分母を因数分解すると、

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\(\begin{eqnarray}~~~k^2+3k+2&=&(k+1)(k+2)
\end{eqnarray}\)

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よって、

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\(\begin{eqnarray}~~~\displaystyle \frac{1}{k^2+3k+2}=\frac{1}{(k+1)(k+2)}\end{eqnarray}\)

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\(\displaystyle \frac{1}{(k+1)(k+2)}\) を部分分数に分けた式は、

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\(\begin{eqnarray}~~~\frac{1}{k+1}-\frac{1}{k+2}&=&\frac{(k+2)-(k+1)}{(k+1)(k+2)}
\\[5pt]~~~&=&\frac{1}{(k+1)(k+2)}
\end{eqnarray}\)

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よって、恒等式

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\(\begin{eqnarray}~~~\displaystyle \frac{1}{(k+1)(k+2)}=\left(\displaystyle \frac{1}{k+1}-\displaystyle \frac{1}{k+2}\right)\end{eqnarray}\)

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が成り立つ

---

これを用いて、

---

\(\begin{eqnarray}~~~&&\sum_{k=1}^{10}\frac{1}{k^2+3k+2}
\\[5pt]~~~&=&\sum_{k=1}^{10}\frac{1}{(k+1)(k+2)}
\\[5pt]~~~&=&\sum_{k=1}^{10}\left(\frac{1}{k+1}-\frac{1}{k+2}\right)
\end{eqnarray}\)

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ここで、\(k=1\sim 10\) までの自然数を代入した和を書き並べると、

---

\(\begin{eqnarray} \require{cancel}~~~=&&\left(\frac{\,1\,}{\,2\,}-\cancel{\frac{\,1\,}{\,3\,}}\right)
\\[5pt]~~~+&&\left(\cancel{\frac{\,1\,}{\,3\,}}-\cancel{\frac{\,1\,}{\,4\,}}\right)
\\[5pt]~~~+&&\left(\cancel{\frac{\,1\,}{\,4\,}}-\cancel{\frac{\,1\,}{\,5\,}}\right)
\\[5pt]~~~&& ~ ~ \cdots
\\[5pt]~~~+&&\left(\cancel{\frac{\,1\,}{\,10\,}}-\cancel{\frac{\,1\,}{\,11\,}}\right)
\\[5pt]~~~+&&\left(\cancel{\frac{\,1\,}{\,11\,}}-\frac{\,1\,}{\,12\,}\right)
\end{eqnarray}\)

---

消えていく項と残る項を確認すると、

---

\(\begin{eqnarray}~~~&=&\frac{\,1\,}{\,2\,}-\frac{\,1\,}{\,12\,}
\\[5pt]~~~&=&\frac{\,6-1\,}{\,12\,}
\\[5pt]~~~&=&\frac{\,5\,}{\,12\,}
\end{eqnarray}\)

東京書籍｜Advanced数学Ｂ[701] p.44 練習問題A 5(2)

分母は自然数の和なので、

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\(\begin{eqnarray}~~~1+2+3+\cdots+k&=&\displaystyle \frac{k(k+1)}{2}
\end{eqnarray}\)

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よって、一般項は、

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\(\begin{eqnarray}~~~\displaystyle \frac{1}{1+2+3+\cdots+k}&=&\frac{1}{\displaystyle \frac{k(k+1)}{2}}
\\[5pt]~~~&=&\frac{2}{k(k+1)}\end{eqnarray}\)

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\(\displaystyle \frac{2}{k(k+1)}\) を部分分数に分けた式は、

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\(\begin{eqnarray}~~~\frac{1}{k}-\frac{1}{k+1}&=&\frac{(k+1)-k}{k(k+1)}
\\[5pt]~~~&=&\frac{1}{k(k+1)}
\end{eqnarray}\)

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※ 分子が \(1\) となるので、両辺を \(2\) 倍する。

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よって、恒等式

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\(\begin{eqnarray}~~~\displaystyle \frac{2}{k(k+1)}=2\left(\displaystyle \frac{1}{k}-\displaystyle \frac{1}{k+1}\right)\end{eqnarray}\)

---

が成り立つ

---

これを用いて、

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\(\begin{eqnarray}~~~&&\sum_{k=1}^{n}\frac{2}{k(k+1)}
\\[5pt]~~~&=&\sum_{k=1}^{n}2\left(\frac{1}{k}-\frac{1}{k+1}\right)
\\[5pt]~~~&=&2\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)
\end{eqnarray}\)

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ここで、\(k=1\sim n\) までの自然数を代入した和を書き並べると、

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\(\begin{eqnarray} \require{cancel}~~~=&&2 \left\{ \left(\frac{\,1\,}{\,1\,}-\cancel{\frac{\,1\,}{\,2\,}}\right)\right.
\\[5pt]~~~&&~ ~+\left(\cancel{\frac{\,1\,}{\,2\,}}-\cancel{\frac{\,1\,}{\,3\,}}\right)
\\[5pt]~~~&&~ ~+\left(\cancel{\frac{\,1\,}{\,3\,}}-\cancel{\frac{\,1\,}{\,4\,}}\right)
\\[5pt]~~~&& ~ ~ ~ ~ ~ ~ ~ ~ \cdots
\\[5pt]~~~&&~ ~+\left(\cancel{\frac{\,1\,}{\,n-1\,}}-\cancel{\frac{\,1\,}{\,n\,}}\right)
\\[5pt]~~~&&~ ~+\left. \left(\cancel{\frac{\,1\,}{\,n\,}}-\frac{\,1\,}{\,n+1\,}\right) \right\}
\end{eqnarray}\)

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消えていく項と残る項を確認すると、

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\(\begin{eqnarray}~~~&=&2 \left(1-\frac{\,1\,}{\,n+1\,}\right)
\\[5pt]~~~&=&2 \cdot \frac{\,(n+1)-1\,}{\,n+1\,}
\\[5pt]~~~&=&2 \cdot \frac{\,n\,}{\,n+1\,}
\\[5pt]~~~&=&\frac{\,2n\,}{\,n+1\,}
\end{eqnarray}\)

東京書籍｜Standard数学Ｂ[702] p.37 問13

与えられた恒等式

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\(\begin{eqnarray}~~~\displaystyle \frac{2}{(2k-1)(2k+1)}=\left(\displaystyle \frac{1}{2k-1}-\displaystyle \frac{1}{2k+1}\right)\end{eqnarray}\)

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を利用する

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\(S_n\) をシグマで表すと、

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\(\begin{eqnarray}~~~S_n&=&\displaystyle \sum_{k=1}^{n}\frac{2}{(2k-1)(2k+1)}
\end{eqnarray}\)

---

これを用いて、

---

\(\begin{eqnarray}~~~&&\sum_{k=1}^{n}\frac{2}{(2k-1)(2k+1)}
\\[5pt]~~~&=&\sum_{k=1}^{n}\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)
\end{eqnarray}\)

---

ここで、\(k=1\sim n\) までの自然数を代入した和を書き並べると、

---

\(\begin{eqnarray} \require{cancel}~~~=&&\left(\frac{\,1\,}{\,1\,}-\cancel{\frac{\,1\,}{\,3\,}}\right)
\\[5pt]~~~+&&\left(\cancel{\frac{\,1\,}{\,3\,}}-\cancel{\frac{\,1\,}{\,5\,}}\right)
\\[5pt]~~~+&&\left(\cancel{\frac{\,1\,}{\,5\,}}-\cancel{\frac{\,1\,}{\,7\,}}\right)
\\[5pt]~~~&& ~ ~ \cdots
\\[5pt]~~~+&&\left(\cancel{\frac{\,1\,}{\,2n-3\,}}-\cancel{\frac{\,1\,}{\,2n-1\,}}\right)
\\[5pt]~~~+&&\left(\cancel{\frac{\,1\,}{\,2n-1\,}}-\frac{\,1\,}{\,2n+1\,}\right)
\end{eqnarray}\)

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消えていく項と残る項を確認すると、

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\(\begin{eqnarray}~~~&=&1-\frac{\,1\,}{\,2n+1\,}
\\[5pt]~~~&=&\frac{\,(2n+1)-1\,}{\,2n+1\,}
\\[5pt]~~~&=&\frac{\,2n\,}{\,2n+1\,}
\end{eqnarray}\)

---

したがって、\(\displaystyle S_n=\frac{\,2n\,}{\,2n+1\,}\)

東京書籍｜Standard数学Ｂ[702] p.40 Training 15

与えられた恒等式

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\(\begin{eqnarray}~~~\displaystyle \frac{1}{(3k-2)(3k+1)}=\frac{1}{3}\left(\displaystyle \frac{1}{3k-2}-\displaystyle \frac{1}{3k+1}\right)\end{eqnarray}\)

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を利用する

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\(S_n\) をシグマで表すと、

---

\(\begin{eqnarray}~~~S_n&=&\displaystyle \sum_{k=1}^{n}\frac{1}{(3k-2)(3k+1)}
\end{eqnarray}\)

---

これを用いて、

---

\(\begin{eqnarray}~~~&&\sum_{k=1}^{n}\frac{1}{(3k-2)(3k+1)}
\\[5pt]~~~&=&\sum_{k=1}^{n}\frac{1}{3}\left(\frac{1}{3k-2}-\frac{1}{3k+1}\right)
\\[5pt]~~~&=&\frac{1}{3}\sum_{k=1}^{n}\left(\frac{1}{3k-2}-\frac{1}{3k+1}\right)
\end{eqnarray}\)

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ここで、\(k=1\sim n\) までの自然数を代入した和を書き並べると、

---

\(\begin{eqnarray} \require{cancel}~~~=&&\frac{1}{3} \left\{ \left(\frac{\,1\,}{\,1\,}-\cancel{\frac{\,1\,}{\,4\,}}\right)\right.
\\[5pt]~~~&&~ ~+\left(\cancel{\frac{\,1\,}{\,4\,}}-\cancel{\frac{\,1\,}{\,7\,}}\right)
\\[5pt]~~~&&~ ~+\left(\cancel{\frac{\,1\,}{\,7\,}}-\cancel{\frac{\,1\,}{\,10\,}}\right)
\\[5pt]~~~&& ~ ~ ~ ~ ~ ~ ~ ~ \cdots
\\[5pt]~~~&&~ ~+\left(\cancel{\frac{\,1\,}{\,3n-5\,}}-\cancel{\frac{\,1\,}{\,3n-2\,}}\right)
\\[5pt]~~~&&~ ~+\left. \left(\cancel{\frac{\,1\,}{\,3n-2\,}}-\frac{\,1\,}{\,3n+1\,}\right) \right\}
\end{eqnarray}\)

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消えていく項と残る項を確認すると、

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\(\begin{eqnarray}~~~&=&\frac{1}{3} \left(1-\frac{\,1\,}{\,3n+1\,}\right)
\\[5pt]~~~&=&\frac{1}{3} \cdot \frac{\,(3n+1)-1\,}{\,3n+1\,}
\\[5pt]~~~&=&\frac{1}{3} \cdot \frac{\,3n\,}{\,3n+1\,}
\\[5pt]~~~&=&\frac{\,n\,}{\,3n+1\,}
\end{eqnarray}\)

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したがって、\(\displaystyle S_n=\frac{\,n\,}{\,3n+1\,}\)