Σ計算(部分分数分解編)
数列(教科書範囲) ★★★
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
$\displaystyle \dfrac{1}{AB}=\dfrac{1}{B-A}\left(\dfrac{1}{A}-\dfrac{1}{B}\right)$
3つパターン
$\displaystyle \dfrac{1}{ABC}=\dfrac{1}{C-A}\left(\dfrac{1}{AB}-\dfrac{1}{BC}\right)$
シグマ計算の部分分数分解は，ナンバリングがズレた差を作るのが目的で，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．
$\displaystyle =\left\{\dfrac{1}{3}\left(\dfrac{1}{2}\right)^{2}-\dfrac{1}{5}\left(\dfrac{1}{2}\right)^{3}\right\}+\left\{\dfrac{1}{5}\left(\dfrac{1}{2}\right)^{3}-\dfrac{1}{7}\left(\dfrac{1}{2}\right)^{4}\right\}+\cdots+\left\{\dfrac{1}{2n+1}\left(\dfrac{1}{2}\right)^{n+1}-\dfrac{1}{2n+3}\left(\dfrac{1}{2}\right)^{n+2}\right\}$
$\displaystyle =\dfrac{1}{3}\left(\dfrac{1}{2}\right)^{2}-\dfrac{1}{2n+3}\left(\dfrac{1}{2}\right)^{n+2}$
$=\boldsymbol{\dfrac{1}{12}-\dfrac{1}{2n+3}\left(\dfrac{1}{2}\right)^{n+2}}$