以下の3つの和を求める問題です。 (1) $\sum_{k=1}^{n} (2k+1)$ (2) $\sum_{k=1}^{n} (3k-5)$ (3) $\sum_{k=1}^{n-1} 4k$代数学数列シグマ和の公式2025/8/61. 問題の内容以下の3つの和を求める問題です。(1) ∑k=1n(2k+1)\sum_{k=1}^{n} (2k+1)∑k=1n(2k+1)(2) ∑k=1n(3k−5)\sum_{k=1}^{n} (3k-5)∑k=1n(3k−5)(3) ∑k=1n−14k\sum_{k=1}^{n-1} 4k∑k=1n−14k2. 解き方の手順(1) ∑k=1n(2k+1)\sum_{k=1}^{n} (2k+1)∑k=1n(2k+1)∑k=1n(2k+1)=2∑k=1nk+∑k=1n1\sum_{k=1}^{n} (2k+1) = 2\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1∑k=1n(2k+1)=2∑k=1nk+∑k=1n1∑k=1nk=n(n+1)2\sum_{k=1}^{n} k = \frac{n(n+1)}{2}∑k=1nk=2n(n+1)∑k=1n1=n\sum_{k=1}^{n} 1 = n∑k=1n1=nよって、∑k=1n(2k+1)=2⋅n(n+1)2+n=n(n+1)+n=n2+n+n=n2+2n=n(n+2)\sum_{k=1}^{n} (2k+1) = 2 \cdot \frac{n(n+1)}{2} + n = n(n+1) + n = n^2 + n + n = n^2 + 2n = n(n+2)∑k=1n(2k+1)=2⋅2n(n+1)+n=n(n+1)+n=n2+n+n=n2+2n=n(n+2)(2) ∑k=1n(3k−5)\sum_{k=1}^{n} (3k-5)∑k=1n(3k−5)∑k=1n(3k−5)=3∑k=1nk−5∑k=1n1\sum_{k=1}^{n} (3k-5) = 3\sum_{k=1}^{n} k - 5\sum_{k=1}^{n} 1∑k=1n(3k−5)=3∑k=1nk−5∑k=1n1∑k=1nk=n(n+1)2\sum_{k=1}^{n} k = \frac{n(n+1)}{2}∑k=1nk=2n(n+1)∑k=1n1=n\sum_{k=1}^{n} 1 = n∑k=1n1=nよって、∑k=1n(3k−5)=3⋅n(n+1)2−5n=3n(n+1)2−10n2=3n2+3n−10n2=3n2−7n2=n(3n−7)2\sum_{k=1}^{n} (3k-5) = 3 \cdot \frac{n(n+1)}{2} - 5n = \frac{3n(n+1)}{2} - \frac{10n}{2} = \frac{3n^2 + 3n - 10n}{2} = \frac{3n^2 - 7n}{2} = \frac{n(3n-7)}{2}∑k=1n(3k−5)=3⋅2n(n+1)−5n=23n(n+1)−210n=23n2+3n−10n=23n2−7n=2n(3n−7)(3) ∑k=1n−14k\sum_{k=1}^{n-1} 4k∑k=1n−14k∑k=1n−14k=4∑k=1n−1k\sum_{k=1}^{n-1} 4k = 4\sum_{k=1}^{n-1} k∑k=1n−14k=4∑k=1n−1k∑k=1n−1k=(n−1)((n−1)+1)2=(n−1)n2\sum_{k=1}^{n-1} k = \frac{(n-1)((n-1)+1)}{2} = \frac{(n-1)n}{2}∑k=1n−1k=2(n−1)((n−1)+1)=2(n−1)nよって、∑k=1n−14k=4⋅(n−1)n2=2n(n−1)=2n2−2n\sum_{k=1}^{n-1} 4k = 4 \cdot \frac{(n-1)n}{2} = 2n(n-1) = 2n^2 - 2n∑k=1n−14k=4⋅2(n−1)n=2n(n−1)=2n2−2n3. 最終的な答え(1) n(n+2)n(n+2)n(n+2)(2) n(3n−7)2\frac{n(3n-7)}{2}2n(3n−7)(3) 2n(n−1)2n(n-1)2n(n−1)
