次の計算をせよ。 (1) $\sum_{k=1}^{6} 1$ (2) $\sum_{k=1}^{10} (3k-2)$ (3) $\sum_{i=1}^{7} 2^i$ (4) $\sum_{i=1}^{n} i^2 = \frac{1}{[ ]}n(n+[ ])([ ]n+[ ])$ (5) $\frac{1}{(3n-2)(3n+1)} = \frac{1}{[ ]}(\frac{1}{3n-2} - \frac{1}{3n+1})$ (6) $\sum_{k=1}^{n} \frac{1}{(3k-2)(3k+1)} = \frac{n}{[ ]n+[ ]}$

はい、承知いたしました。画像にある数学の問題を解いていきます。

1. 問題の内容

次の計算をせよ。

(1) $\sum_{k=1}^{6} 1$

(2) $\sum_{k=1}^{10} (3k-2)$

(3) $\sum_{i=1}^{7} 2^i$

(4) $\sum_{i=1}^{n} i^2 = \frac{1}{[ ]}n(n+[ ])([ ]n+[ ])$

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

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

2. 解き方の手順

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

(2) $\sum_{k=1}^{10} (3k-2) = 3\sum_{k=1}^{10} k - \sum_{k=1}^{10} 2 = 3\frac{10(10+1)}{2} - 2(10) = 3\frac{10(11)}{2} - 20 = 3(55) - 20 = 165 - 20 = 145$

(3) $\sum_{i=1}^{7} 2^i = 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 = 2\frac{2^7-1}{2-1} = 2(128-1) = 2(127) = 254$

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

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

$1 = A(3n+1) + B(3n-2)$

$n = -\frac{1}{3}$のとき: $1 = B(3(-\frac{1}{3})-2) = B(-1-2) = -3B$ よって $B = -\frac{1}{3}$

$n = \frac{2}{3}$のとき: $1 = A(3(\frac{2}{3})+1) = A(2+1) = 3A$ よって $A = \frac{1}{3}$

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

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

$= \frac{1}{3}[(\frac{1}{1} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{7}) + (\frac{1}{7} - \frac{1}{10}) + ... + (\frac{1}{3n-2} - \frac{1}{3n+1})] = \frac{1}{3}(1 - \frac{1}{3n+1}) = \frac{1}{3}(\frac{3n+1-1}{3n+1}) = \frac{1}{3}(\frac{3n}{3n+1}) = \frac{n}{3n+1}$

3. 最終的な答え

(1) 6

(2) 145

(3) 254

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

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

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