与えられた級数の和を求めます。具体的には、$\sum_{k=1}^{n} \frac{3+2k}{k(k+1)(k+2)(k+3)}$ を計算します。

1. 問題の内容

与えられた級数の和を求めます。具体的には、$\sum_{k=1}^{n} \frac{3+2k}{k(k+1)(k+2)(k+3)}$ を計算します。

2. 解き方の手順

部分分数分解を利用して、与えられた項をより扱いやすい形に変形します。

$\frac{3+2k}{k(k+1)(k+2)(k+3)} = \frac{A}{k} + \frac{B}{k+1} + \frac{C}{k+2} + \frac{D}{k+3}$

両辺に $k(k+1)(k+2)(k+3)$ をかけると、

$3+2k = A(k+1)(k+2)(k+3) + B k(k+2)(k+3) + C k(k+1)(k+3) + D k(k+1)(k+2)$

$k=0$ のとき、$3 = A(1)(2)(3) \implies A = \frac{1}{2}$

$k=-1$ のとき、$1 = B(-1)(1)(2) \implies B = -\frac{1}{2}$

$k=-2$ のとき、$-1 = C(-2)(-1)(1) \implies C = -\frac{1}{2}$

$k=-3$ のとき、$-3 = D(-3)(-2)(-1) \implies D = \frac{1}{2}$

したがって、

$\frac{3+2k}{k(k+1)(k+2)(k+3)} = \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+1} - \frac{1}{k+2} + \frac{1}{k+3} \right)$

ここで、

$\frac{1}{k} - \frac{1}{k+1} = \frac{1}{k(k+1)}$

$\frac{1}{k+2} - \frac{1}{k+3} = \frac{1}{(k+2)(k+3)}$

したがって、

$\frac{1}{k} - \frac{1}{k+1} - \frac{1}{k+2} + \frac{1}{k+3} = \frac{1}{k(k+1)} - \frac{1}{(k+2)(k+3)}$

$= \frac{(k+2)(k+3) - k(k+1)}{k(k+1)(k+2)(k+3)} = \frac{k^2+5k+6-k^2-k}{k(k+1)(k+2)(k+3)} = \frac{4k+6}{k(k+1)(k+2)(k+3)}$

別の変形を試みます。

$\frac{1}{k} - \frac{1}{k+1} - \frac{1}{k+2} + \frac{1}{k+3} = \left( \frac{1}{k} - \frac{1}{k+1} \right) - \left( \frac{1}{k+2} - \frac{1}{k+3} \right)$

$=\frac{1}{k(k+1)} - \frac{1}{(k+2)(k+3)} = \frac{(k+2)(k+3) - k(k+1)}{k(k+1)(k+2)(k+3)}$

$=\frac{k^2+5k+6-k^2-k}{k(k+1)(k+2)(k+3)} = \frac{4k+6}{k(k+1)(k+2)(k+3)}$

この変形は元に戻ってしまっています。

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

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

= \frac{1}{2} \left[ \left( 1 - \frac{1}{n+1} \right) - \left( \frac{1}{3} - \frac{1}{n+3} \right) - \left( \frac{1}{2} - \frac{1}{n+2} \right) + \left( \frac{1}{3}-\frac{1}{4} \right) + \left(\frac{1}{4}-\frac{1}{5}\right) + \dots

$= \frac{1}{2} \left[ \left( 1+\frac{1}{2} \right) - \left(\frac{1}{n+1}+\frac{1}{n+2}\right) - \left(\frac{1}{2} - \frac{1}{n+3} \right) - \left(\frac{1}{3} - \frac{1}{n+3} \right) \right]$

$= \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{3} - \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} \right)$

$= \frac{1}{2} \left( \frac{7}{6} - \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} \right)$

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

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

$= (1-\frac{1}{n+1}) - (\frac{1}{3}-\frac{1}{n+3} + \frac{1}{4}-\frac{1}{n+4}...) = 1-\frac{1}{2}-\frac{1}{3} + (\frac{1}{n+1}-\frac{1}{n+3})$

$= 1+\frac{1}{2} - (\frac{1}{n+1} + \frac{1}{n+2}) +\frac{1}{2}( (\frac{1}{n+3} + \frac{1}{n+4})$.

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

$= (1-\frac{1}{2} + \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+ ... + \frac{1}{n} - \frac{1}{n+1}) - (\frac{1}{3}-\frac{1}{4} + \frac{1}{4}-\frac{1}{5} + ... + \frac{1}{n+2}-\frac{1}{n+3})$

$= (1-\frac{1}{n+1}) - (\frac{1}{3} - \frac{1}{n+3})$

$= 1-\frac{1}{3} -\frac{1}{n+1} + \frac{1}{n+3}$

$= \frac{2}{3} - \frac{1}{n+1} + \frac{1}{n+3} = \frac{2}{3} - \frac{(n+3)-(n+1)}{(n+1)(n+3)} = \frac{2}{3} - \frac{2}{(n+1)(n+3)}$

$\frac{1}{2} \left( \frac{2}{3} - \frac{2}{(n+1)(n+3)} \right) = \frac{1}{3} - \frac{1}{(n+1)(n+3)}$

3. 最終的な答え

$\frac{1}{3} - \frac{1}{(n+1)(n+3)}$