高専の数学解説

第1章 1.20

(1)

∑k=1n(2k−1)(2k+2)=∑k=1n(4k2+2k−2)=4∑k=1nk2+2∑k=1nk−2∑k=1n1\sum_{k=1}^{n}(2k-1)(2k+2) = \sum_{k=1}^{n}(4k^2+2k-2) = 4\sum_{k=1}^{n}k^2 + 2\sum_{k=1}^{n}k - 2\sum_{k=1}^{n}1

=4⋅n(n+1)(2n+1)6+2⋅n(n+1)2−2n=13n{2(n+1)(2n+1+32)−6}= 4\cdot\frac{n(n+1)(2n+1)}{6} + 2\cdot\frac{n(n+1)}{2} - 2n = \frac{1}{3}n\left\{2(n+1)\left(2n+1+\frac{3}{2}\right) - 6\right\}

=13n(4n2+5n+4n+5−6)=n(4n2+9n−1)3= \frac{1}{3}n(4n^2+5n+4n+5-6) = \frac{n(4n^2+9n-1)}{3}

(2)

∑k=1n∑i=1ki=∑k=1nk(k+1)2=12∑k=1nk2+12∑k=1nk=12⋅n(n+1)(2n+1)6+12⋅n(n+1)2\sum_{k=1}^{n}\sum_{i=1}^{k} i = \sum_{k=1}^{n}\frac{k(k+1)}{2} = \frac{1}{2}\sum_{k=1}^{n}k^2 + \frac{1}{2}\sum_{k=1}^{n}k = \frac{1}{2}\cdot\frac{n(n+1)(2n+1)}{6} + \frac{1}{2}\cdot\frac{n(n+1)}{2}

=112⋅n(n+1)(2n+1+3)=16⋅n(n+1)(n+2)= \frac{1}{12}\cdot n(n+1)(2n+1+3) = \frac{1}{6}\cdot n(n+1)(n+2)

(3)

∑k=1n(2k−1)⋅2k⋅(2k+1)=8∑k=1nk3−2∑k=1nk=8⋅n2(n+1)24−2⋅n(n+1)2\sum_{k=1}^{n}(2k-1)\cdot 2k \cdot (2k+1) = 8\sum_{k=1}^{n}k^3 - 2\sum_{k=1}^{n}k = 8\cdot\frac{n^2(n+1)^2}{4} - 2\cdot\frac{n(n+1)}{2}

=n(n+1){2n(n+1)−1}=n(n+1)(2n2+2n−1)= n(n+1)\{2n(n+1)-1\} = n(n+1)(2n^2+2n-1)

解説: r31bn1z