以下の6つの和を求める問題です。 (1) $\sum_{k=1}^{n} (2k - 1)$ (2) $\sum_{k=1}^{n} (6k^2 - 4k + 5)$ (3) $\sum_{k=1}^{n} (4k^3 - 6k)$ (4) $\sum_{k=1}^{n} 2^k$ (5) $\sum_{k=1}^{n-1} (12k^2 + 4k - 1)$ (6) $\sum_{k=1}^{n+1} 3^{k-1}$

1. 問題の内容

以下の6つの和を求める問題です。

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

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

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

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

(5) $\sum_{k=1}^{n-1} (12k^2 + 4k - 1)$

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

2. 解き方の手順

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

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

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

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

$= 6\frac{n(n+1)(2n+1)}{6} - 4\frac{n(n+1)}{2} + 5n = n(n+1)(2n+1) - 2n(n+1) + 5n$

$= n( (n+1)(2n+1) - 2(n+1) + 5) = n(2n^2 + 3n + 1 - 2n - 2 + 5) = n(2n^2 + n + 4) = 2n^3 + n^2 + 4n$

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

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

$= n(n+1) (n(n+1) - 3) = n(n+1)(n^2 + n - 3) = n(n^3 + n^2 - 3n + n^2 + n - 3) = n(n^3 + 2n^2 - 2n - 3) = n^4 + 2n^3 - 2n^2 - 3n$

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

等比数列の和の公式を使用します。$\sum_{k=1}^{n} ar^{k-1} = a\frac{1-r^n}{1-r}$

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

(5) $\sum_{k=1}^{n-1} (12k^2 + 4k - 1)$

$\sum_{k=1}^{n-1} (12k^2 + 4k - 1) = 12\sum_{k=1}^{n-1} k^2 + 4\sum_{k=1}^{n-1} k - \sum_{k=1}^{n-1} 1 = 12\frac{(n-1)n(2(n-1)+1)}{6} + 4\frac{(n-1)n}{2} - (n-1)$

$= 2(n-1)n(2n-1) + 2(n-1)n - (n-1) = (n-1)(2n(2n-1) + 2n - 1) = (n-1)(4n^2 - 2n + 2n - 1) = (n-1)(4n^2 - 1)$

$= 4n^3 - n - 4n^2 + 1 = 4n^3 - 4n^2 - n + 1$

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

等比数列の和の公式を使用します。$\sum_{k=1}^{m} ar^{k-1} = a\frac{1-r^m}{1-r}$

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

3. 最終的な答え

(1) $n^2$

(2) $2n^3 + n^2 + 4n$

(3) $n^4 + 2n^3 - 2n^2 - 3n$

(4) $2^{n+1} - 2$

(5) $4n^3 - 4n^2 - n + 1$

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