(1) 次の和を求めよ。 (i) $\sum_{k=1}^{n} (4k+3) = \boxed{1} n^2 + \boxed{2} n$ (ii) $\sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)} = \frac{n}{\boxed{3} n + \boxed{4}}$

1. 問題の内容

(1) 次の和を求めよ。

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

(ii) $\sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)} = \frac{n}{\boxed{3} n + \boxed{4}}$

2. 解き方の手順

(i) $\sum_{k=1}^{n} (4k+3)$を計算する。

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

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

$= 2n(n+1) + 3n$

$= 2n^2 + 2n + 3n$

$= 2n^2 + 5n$

よって、$2n^2 + 5n = \boxed{2}n^2 + \boxed{5}n$

(ii) $\sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)}$を計算する。

$\frac{1}{(2k-1)(2k+1)} = \frac{A}{2k-1} + \frac{B}{2k+1}$とおく。

$1 = A(2k+1) + B(2k-1)$

$1 = (2A+2B)k + (A-B)$

よって、$2A+2B = 0$かつ$A-B=1$

$A+B=0$かつ$A-B=1$

$2A=1$より$A=\frac{1}{2}$

$B = -A = -\frac{1}{2}$

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

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

$= \frac{1}{2}[(\frac{1}{1} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{5}) + \dots + (\frac{1}{2n-1} - \frac{1}{2n+1})]$

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

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

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

$= \frac{n}{2n+1}$

$2n+1 = \boxed{2}n + \boxed{1}$

3. 最終的な答え

(i) $\boxed{2} n^2 + \boxed{5} n$

(ii) $\frac{n}{\boxed{2} n + \boxed{1}}$