数列 $a_i$ と $b_i$ が与えられたとき、以下の総和を計算する問題です。 (1) $\sum_{i=1}^{7} a_i$ (2) $\sum_{i=2}^{6} a_i$ (3) $\sum_{i=1}^{9} (a_i + 3)$ (4) $\sum_{i=1}^{9} a_i^2$ (5) $\sum_{i=1}^{6} b_i$ (6) $\sum_{i=1}^{9} a_i b_i$ (7) $\sum_{i=1}^{9} (a_i + b_i)$ (8) $\sum_{i=1}^{8} (a_i + b_i)^2$

1. 問題の内容

数列 $a_i$ と $b_i$ が与えられたとき、以下の総和を計算する問題です。

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

(2) $\sum_{i=2}^{6} a_i$

(3) $\sum_{i=1}^{9} (a_i + 3)$

(4) $\sum_{i=1}^{9} a_i^2$

(5) $\sum_{i=1}^{6} b_i$

(6) $\sum_{i=1}^{9} a_i b_i$

(7) $\sum_{i=1}^{9} (a_i + b_i)$

(8) $\sum_{i=1}^{8} (a_i + b_i)^2$

2. 解き方の手順

(1) $\sum_{i=1}^{7} a_i = a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 = 4 + 6 + 1 + 7 + 8 + 5 + 3$

(2) $\sum_{i=2}^{6} a_i = a_2 + a_3 + a_4 + a_5 + a_6 = 6 + 1 + 7 + 8 + 5$

(3) $\sum_{i=1}^{9} (a_i + 3) = (a_1 + 3) + (a_2 + 3) + ... + (a_9 + 3) = \sum_{i=1}^{9} a_i + \sum_{i=1}^{9} 3 = (a_1 + a_2 + ... + a_9) + 9 \times 3$

(4) $\sum_{i=1}^{9} a_i^2 = a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 + a_6^2 + a_7^2 + a_8^2 + a_9^2$

(5) $\sum_{i=1}^{6} b_i = b_1 + b_2 + b_3 + b_4 + b_5 + b_6 = 1 + 3 + 2 + 4 + 5 + 2$

(6) $\sum_{i=1}^{9} a_i b_i = a_1 b_1 + a_2 b_2 + a_3 b_3 + a_4 b_4 + a_5 b_5 + a_6 b_6 + a_7 b_7 + a_8 b_8 + a_9 b_9$

(7) $\sum_{i=1}^{9} (a_i + b_i) = (a_1 + b_1) + (a_2 + b_2) + ... + (a_9 + b_9) = \sum_{i=1}^{9} a_i + \sum_{i=1}^{9} b_i$

(8) $\sum_{i=1}^{8} (a_i + b_i)^2 = (a_1 + b_1)^2 + (a_2 + b_2)^2 + ... + (a_8 + b_8)^2$

各数列の値を代入して計算します。

(1) $\sum_{i=1}^{7} a_i = 4 + 6 + 1 + 7 + 8 + 5 + 3 = 34$

(2) $\sum_{i=2}^{6} a_i = 6 + 1 + 7 + 8 + 5 = 27$

(3) $\sum_{i=1}^{9} a_i = 4 + 6 + 1 + 7 + 8 + 5 + 3 + 6 + 2 = 42$

$\sum_{i=1}^{9} (a_i + 3) = 42 + 9 \times 3 = 42 + 27 = 69$

(4) $\sum_{i=1}^{9} a_i^2 = 4^2 + 6^2 + 1^2 + 7^2 + 8^2 + 5^2 + 3^2 + 6^2 + 2^2 = 16 + 36 + 1 + 49 + 64 + 25 + 9 + 36 + 4 = 240$

(5) $\sum_{i=1}^{6} b_i = 1 + 3 + 2 + 4 + 5 + 2 = 17$

(6) $\sum_{i=1}^{9} a_i b_i = (4 \times 1) + (6 \times 3) + (1 \times 2) + (7 \times 4) + (8 \times 5) + (5 \times 2) + (3 \times 4) + (6 \times 1) + (2 \times 6) = 4 + 18 + 2 + 28 + 40 + 10 + 12 + 6 + 12 = 132$

(7) $\sum_{i=1}^{9} b_i = 1 + 3 + 2 + 4 + 5 + 2 + 4 + 1 + 6 = 28$

$\sum_{i=1}^{9} (a_i + b_i) = 42 + 28 = 70$

(8) $\sum_{i=1}^{8} (a_i + b_i)^2 = (4+1)^2 + (6+3)^2 + (1+2)^2 + (7+4)^2 + (8+5)^2 + (5+2)^2 + (3+4)^2 + (6+1)^2 = 5^2 + 9^2 + 3^2 + 11^2 + 13^2 + 7^2 + 7^2 + 7^2 = 25 + 81 + 9 + 121 + 169 + 49 + 49 + 49 = 551$

3. 最終的な答え

(1) 34

(2) 27

(3) 69

(4) 240

(5) 17

(6) 132

(7) 70

(8) 551