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Σ計算(部分分数分解編)

数列(教科書範囲) ★★★


アイキャッチ

今回は,Σ公式が使えない和に対して,どうアプローチするかという話です.

数学Ⅱで学習した部分分数分解等を使って解く問題を扱います.



数列でよく使う部分分数分解

ポイント

数列でよく使う部分分数分解

2つパターン

$\displaystyle \dfrac{1}{AB}=\dfrac{1}{B-A}\left(\dfrac{1}{A}-\dfrac{1}{B}\right)$

※ $A$,$B$ は多項式で,$B-A$ が定数

3つパターン

$\displaystyle \dfrac{1}{ABC}=\dfrac{1}{C-A}\left(\dfrac{1}{AB}-\dfrac{1}{BC}\right)$

※ $A$,$B$,$C$は多項式で,$C-A$ が定数


機械的に上の式に当てはめるというより,ナンバリングがズレた差を作るのが目的です.すべてバラバラにする必要はありません.

証明は右辺展開で簡単です.下の問題で使い方を確認します.

例題と練習問題

例題

例題

次の計算をせよ.

(1) $\displaystyle \sum_{k=1}^{n}\dfrac{1}{k(k+1)}$

(2) $\displaystyle \sum_{k=1}^{n}\dfrac{1}{\sqrt{k+1}+\sqrt{k}}$


講義

$\boldsymbol{k}$ が分母にあるのでΣ公式は使えません.

部分分数分解公式使用か有理化等をして,強引にナンバリングがずれた差を作ります.


解答

(1)

 $\displaystyle \sum_{k=1}^{n}\dfrac{1}{k(k+1)}$

$\displaystyle =\sum_{k=1}^{n}$$\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)$ ←部分分数分解

$\displaystyle =\left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\cdots+\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)+\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)$

↑書き出します.なんと項の隣同士が消えて,最初と最後が残ります

$\displaystyle =\dfrac{1}{1}-\dfrac{1}{n+1}$

$=\boldsymbol{\dfrac{n}{n+1}}$


(2) この問題は部分分数分解は使いませんが,隣同士を消すという考え方は同じです.

 $\displaystyle \sum_{k=1}^{n}\dfrac{1}{\sqrt{k+1}+\sqrt{k}}$$\cdot \dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k+1}-\sqrt{k}}$

$\displaystyle =\sum_{k=1}^{n}\left(-\sqrt{k}+\sqrt{k+1}\right)$

$=(-1+\sqrt{2})+(-\sqrt{2}+\sqrt{3})+(-\sqrt{3}+\sqrt{4})+\cdots+(-\sqrt{n-1}+\sqrt{n})+(-\sqrt{n}+\sqrt{n+1})$

↑書き出します.なんと項の隣同士が消えて,最初と最後が残ります

$=\boldsymbol{-1+\sqrt{n+1}}$

練習問題

練習

次の計算をせよ.

(1) $\displaystyle \dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\cdots+\dfrac{1}{99\cdot101}$

(2) $\displaystyle \sum_{k=1}^{n}\dfrac{1}{k(k+2)}$

(3) $\displaystyle \sum_{k=1}^{n}\dfrac{1}{\sqrt{2k-1}+\sqrt{2k+1}}$

(4) $\displaystyle \sum_{k=1}^{n}\dfrac{1}{(k+1)(k+2)(k+3)}$

(5) $\displaystyle \sum_{k=1}^{n}\dfrac{2k+5}{(2k+1)(2k+3)}\left(\dfrac{1}{2}\right)^{k+2}$

解答

(1) $\displaystyle \dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\cdots+\dfrac{1}{99\cdot101}$

$\displaystyle =\sum_{k=1}^{50}\dfrac{1}{(2k-1)(2k+1)}$

$\displaystyle =\sum_{k=1}^{50}\dfrac{1}{2k+1-(2k-1)}\left(\dfrac{1}{2k-1}-\dfrac{1}{2k+1}\right)$

$\displaystyle =\dfrac{1}{2}\sum_{k=1}^{50}\left(\dfrac{1}{2k-1}-\dfrac{1}{2k+1}\right)$

$\displaystyle =\dfrac{1}{2}\left\{\left(\dfrac{1}{1}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\cdots+\left(\dfrac{1}{97}-\dfrac{1}{99}\right)+\left(\dfrac{1}{99}-\dfrac{1}{101}\right)\right\}$

$\displaystyle =\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)$

$=\boldsymbol{\dfrac{50}{101}}$


(2) $\displaystyle \sum_{k=1}^{n}\dfrac{1}{k(k+2)}$

$\displaystyle =\sum_{k=1}^{n}\dfrac{1}{k+2-k}\left(\dfrac{1}{k}-\dfrac{1}{k+2}\right)$

$\displaystyle =\dfrac{1}{2}\sum_{k=1}^{n}\left(\dfrac{1}{k}-\dfrac{1}{k+2}\right)$

$\displaystyle =\dfrac{1}{2}\left\{\left(\dfrac{1}{1}-\dfrac{1}{3}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\cdots+\left(\dfrac{1}{n-2}-\dfrac{1}{n}\right)+\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)+\left(\dfrac{1}{n}-\dfrac{1}{n+2}\right)\right\}$

↑第 $1$ 項と第 $2$ 項の左,第 $n-1$ 項と第 $n$ 項の右が残ります.

$\displaystyle =\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n+1}-\dfrac{1}{n+2}\right)$

$=\boldsymbol{\dfrac{n(3n+5)}{4(n+1)(n+2)}}$


(3) $\displaystyle \sum_{k=1}^{n}\dfrac{1}{\sqrt{2k-1}+\sqrt{2k+1}}\dfrac{\sqrt{2k-1}-\sqrt{2k+1}}{\sqrt{2k-1}-\sqrt{2k+1}}$

$\displaystyle =\dfrac{1}{2}\sum_{k=1}^{n}(-\sqrt{2k-1}+\sqrt{2k+1})$

$\displaystyle =\dfrac{1}{2}\left\{\left(-1+\sqrt{3}\right)+\left(-\sqrt{3}+\sqrt{5}\right)+\cdots+\left(-\sqrt{2n-3}+\sqrt{2n-1}\right)+\left(-\sqrt{2n-1}+\sqrt{2n+1}\right)\right\}$

$=\boldsymbol{\dfrac{1}{2}\left(-1+\sqrt{2n+1}\right)}$


(4) 3つバージョンですが,部分分数分解にあるように3つバラバラにはせず,上の公式を使います.

 $\displaystyle \sum_{k=1}^{n}\dfrac{1}{(k+1)(k+2)(k+3)}$

$\displaystyle =\sum_{k=1}^{n}\dfrac{1}{k+3-(k+1)}\left\{\dfrac{1}{(k+1)(k+2)}-\dfrac{1}{(k+2)(k+3)}\right\}$

 ↑$\displaystyle \dfrac{1}{ABC}=\dfrac{1}{C-A}\left(\dfrac{1}{AB}-\dfrac{1}{BC}\right)$

$\displaystyle =\dfrac{1}{2}\sum_{k=1}^{n}\left\{\dfrac{1}{(k+1)(k+2)}-\dfrac{1}{(k+2)(k+3)}\right\}$

$\displaystyle =\dfrac{1}{2}\left\{\left(\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}\right)+\left(\dfrac{1}{3\cdot4}-\dfrac{1}{4\cdot5}\right)+\cdots+\left(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right)+\left(\dfrac{1}{(n+1)(n+2)}-\dfrac{1}{(n+2)(n+3)}\right)\right\}$

$\displaystyle =\dfrac{1}{2}\left\{\dfrac{1}{2\cdot3}-\dfrac{1}{(n+2)(n+3)}\right\}$

$=\boldsymbol{\dfrac{n(n+5)}{12(n+2)(n+3)}}$


(5) 2018国際医療福祉大の問題の一部です.

 $\displaystyle \sum_{k=1}^{n}\dfrac{2k+5}{(2k+1)(2k+3)}\left(\dfrac{1}{2}\right)^{k+2}$

$\displaystyle =\sum_{k=1}^{n}\left(\dfrac{2}{2k+1}-\dfrac{1}{2k+3}\right)\left(\dfrac{1}{2}\right)^{k+2}$

$\displaystyle =\sum_{k=1}^{n}\left\{\dfrac{1}{2k+1}\left(\dfrac{1}{2}\right)^{k+1}-\dfrac{1}{2k+3}\left(\dfrac{1}{2}\right)^{k+2}\right\}$

↑強引にナンバリングがずれた差を作ります.

$\displaystyle =\left\{\dfrac{1}{3}\left(\dfrac{1}{2}\right)^{2}-\dfrac{1}{5}\left(\dfrac{1}{2}\right)^{3}\right\}+\left\{\dfrac{1}{5}\left(\dfrac{1}{2}\right)^{3}-\dfrac{1}{7}\left(\dfrac{1}{2}\right)^{4}\right\}+\cdots+\left\{\dfrac{1}{2n+1}\left(\dfrac{1}{2}\right)^{n+1}-\dfrac{1}{2n+3}\left(\dfrac{1}{2}\right)^{n+2}\right\}$

$\displaystyle =\dfrac{1}{3}\left(\dfrac{1}{2}\right)^{2}-\dfrac{1}{2n+3}\left(\dfrac{1}{2}\right)^{n+2}$

$=\boldsymbol{\dfrac{1}{12}-\dfrac{1}{2n+3}\left(\dfrac{1}{2}\right)^{n+2}}$