次の6つの和の計算問題を解きます。 (1) $\sum_{k=1}^{5} 2k$ (2) $\sum_{k=1}^{5} \frac{1}{5}k^2$ (3) $\sum_{t=1}^{5} (t^2 - 5t - 6)$ (4) $\sum_{k=1}^{4} (3^k - 2^k)$ (5) $\sum_{i=1}^{4} \sum_{j=1}^{4} (2i - j)$ (6) $\sum_{i=1}^{4} \sum_{j=1}^{i} (2i - j)$

1. 問題の内容

次の6つの和の計算問題を解きます。

(1) $\sum_{k=1}^{5} 2k$

(2) $\sum_{k=1}^{5} \frac{1}{5}k^2$

(3) $\sum_{t=1}^{5} (t^2 - 5t - 6)$

(4) $\sum_{k=1}^{4} (3^k - 2^k)$

(5) $\sum_{i=1}^{4} \sum_{j=1}^{4} (2i - j)$

(6) $\sum_{i=1}^{4} \sum_{j=1}^{i} (2i - j)$

2. 解き方の手順

(1) $\sum_{k=1}^{5} 2k = 2\sum_{k=1}^{5} k = 2 \cdot \frac{5(5+1)}{2} = 5 \cdot 6 = 30$

(2) $\sum_{k=1}^{5} \frac{1}{5}k^2 = \frac{1}{5} \sum_{k=1}^{5} k^2 = \frac{1}{5} \cdot \frac{5(5+1)(2\cdot 5 + 1)}{6} = \frac{1}{5} \cdot \frac{5 \cdot 6 \cdot 11}{6} = 11$

(3) $\sum_{t=1}^{5} (t^2 - 5t - 6) = \sum_{t=1}^{5} t^2 - 5\sum_{t=1}^{5} t - \sum_{t=1}^{5} 6$

$= \frac{5(5+1)(2\cdot 5 + 1)}{6} - 5 \cdot \frac{5(5+1)}{2} - 6 \cdot 5 = \frac{5 \cdot 6 \cdot 11}{6} - \frac{5 \cdot 5 \cdot 6}{2} - 30 = 55 - 75 - 30 = -50$

(4) $\sum_{k=1}^{4} (3^k - 2^k) = \sum_{k=1}^{4} 3^k - \sum_{k=1}^{4} 2^k = \frac{3(3^4 - 1)}{3-1} - \frac{2(2^4 - 1)}{2-1} = \frac{3(81 - 1)}{2} - 2(16-1) = \frac{3 \cdot 80}{2} - 2 \cdot 15 = 120 - 30 = 90$

(5) $\sum_{i=1}^{4} \sum_{j=1}^{4} (2i - j) = \sum_{i=1}^{4} \left( \sum_{j=1}^{4} 2i - \sum_{j=1}^{4} j \right) = \sum_{i=1}^{4} \left( 2i \cdot 4 - \frac{4(4+1)}{2} \right) = \sum_{i=1}^{4} (8i - 10) = 8 \sum_{i=1}^{4} i - \sum_{i=1}^{4} 10 = 8 \cdot \frac{4(4+1)}{2} - 10 \cdot 4 = 8 \cdot 10 - 40 = 80 - 40 = 40$

(6) $\sum_{i=1}^{4} \sum_{j=1}^{i} (2i - j) = \sum_{i=1}^{4} \left( \sum_{j=1}^{i} 2i - \sum_{j=1}^{i} j \right) = \sum_{i=1}^{4} \left( 2i \cdot i - \frac{i(i+1)}{2} \right) = \sum_{i=1}^{4} \left( 2i^2 - \frac{i^2 + i}{2} \right) = \sum_{i=1}^{4} \left( \frac{4i^2 - i^2 - i}{2} \right) = \sum_{i=1}^{4} \left( \frac{3i^2 - i}{2} \right) = \frac{1}{2} \sum_{i=1}^{4} (3i^2 - i) = \frac{1}{2} \left( 3\sum_{i=1}^{4} i^2 - \sum_{i=1}^{4} i \right) = \frac{1}{2} \left( 3 \cdot \frac{4(4+1)(2\cdot 4 + 1)}{6} - \frac{4(4+1)}{2} \right) = \frac{1}{2} \left( 3 \cdot \frac{4 \cdot 5 \cdot 9}{6} - \frac{4 \cdot 5}{2} \right) = \frac{1}{2} (3 \cdot 30 - 10) = \frac{1}{2} (90 - 10) = \frac{1}{2} \cdot 80 = 40$

3. 最終的な答え

(1) 30

(2) 11

(3) -50

(4) 90

(5) 40

(6) 40