与えられた3つの和を計算します。 (1) $\sum_{k=1}^{n} k(k+1)$ (2) $\sum_{k=1}^{n} n \cdot 2^{k-1}$ (3) $1\cdot 2 + 3\cdot 4 + 5\cdot 6 + \cdots + (2n-1)\cdot 2n$

1. 問題の内容

与えられた3つの和を計算します。

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

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

(3) $1\cdot 2 + 3\cdot 4 + 5\cdot 6 + \cdots + (2n-1)\cdot 2n$

2. 解き方の手順

(1)

まず、$k(k+1)$ を展開します。

$k(k+1) = k^2 + k$

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

$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

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

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

(2)

$\sum_{k=1}^{n} n \cdot 2^{k-1} = n \sum_{k=1}^{n} 2^{k-1}$

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

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

(3)

一般項を求めます。$a_n = (2n-1)(2n) = 4n^2 - 2n$

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

3. 最終的な答え

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

(2) $n(2^n - 1)$

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