次の和を求めます。 $1\cdot2\cdot3 + 2\cdot3\cdot5 + 3\cdot4\cdot7 + \dots + n(n+1)(2n+1)$代数学級数シグマ和の公式多項式2025/8/151. 問題の内容次の和を求めます。1⋅2⋅3+2⋅3⋅5+3⋅4⋅7+⋯+n(n+1)(2n+1)1\cdot2\cdot3 + 2\cdot3\cdot5 + 3\cdot4\cdot7 + \dots + n(n+1)(2n+1)1⋅2⋅3+2⋅3⋅5+3⋅4⋅7+⋯+n(n+1)(2n+1)2. 解き方の手順一般項 aka_kak は次のようになります。ak=k(k+1)(2k+1)a_k = k(k+1)(2k+1)ak=k(k+1)(2k+1)したがって、求める和は∑k=1nk(k+1)(2k+1)=∑k=1nk(2k2+3k+1)=∑k=1n(2k3+3k2+k)\sum_{k=1}^{n} k(k+1)(2k+1) = \sum_{k=1}^{n} k(2k^2+3k+1) = \sum_{k=1}^{n} (2k^3 + 3k^2 + k)∑k=1nk(k+1)(2k+1)=∑k=1nk(2k2+3k+1)=∑k=1n(2k3+3k2+k)和の公式を用いて計算します。∑k=1nk=n(n+1)2\sum_{k=1}^{n} k = \frac{n(n+1)}{2}∑k=1nk=2n(n+1)∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}∑k=1nk2=6n(n+1)(2n+1)∑k=1nk3=(n(n+1)2)2\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2∑k=1nk3=(2n(n+1))2したがって、∑k=1n(2k3+3k2+k)=2∑k=1nk3+3∑k=1nk2+∑k=1nk\sum_{k=1}^{n} (2k^3 + 3k^2 + k) = 2\sum_{k=1}^{n} k^3 + 3\sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k∑k=1n(2k3+3k2+k)=2∑k=1nk3+3∑k=1nk2+∑k=1nk=2(n(n+1)2)2+3n(n+1)(2n+1)6+n(n+1)2= 2 \left(\frac{n(n+1)}{2}\right)^2 + 3 \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}=2(2n(n+1))2+36n(n+1)(2n+1)+2n(n+1)=n2(n+1)22+n(n+1)(2n+1)2+n(n+1)2= \frac{n^2(n+1)^2}{2} + \frac{n(n+1)(2n+1)}{2} + \frac{n(n+1)}{2}=2n2(n+1)2+2n(n+1)(2n+1)+2n(n+1)=n(n+1)2[n(n+1)+(2n+1)+1]= \frac{n(n+1)}{2} [n(n+1) + (2n+1) + 1]=2n(n+1)[n(n+1)+(2n+1)+1]=n(n+1)2[n2+n+2n+2]= \frac{n(n+1)}{2} [n^2 + n + 2n + 2]=2n(n+1)[n2+n+2n+2]=n(n+1)2[n2+3n+2]= \frac{n(n+1)}{2} [n^2 + 3n + 2]=2n(n+1)[n2+3n+2]=n(n+1)(n+1)(n+2)2= \frac{n(n+1)(n+1)(n+2)}{2}=2n(n+1)(n+1)(n+2)=n(n+1)2(n+2)2= \frac{n(n+1)^2(n+2)}{2}=2n(n+1)2(n+2)3. 最終的な答えn(n+1)2(n+2)2\frac{n(n+1)^2(n+2)}{2}2n(n+1)2(n+2)
