高専の数学解説

第1章 1.11

(1)

∑k=1n(2k)2=4∑k=1nk2=4⋅n(n+1)(2n+1)6=2n(n+1)(2n+1)3\sum_{k=1}^n (2k)^2 = 4\sum_{k=1}^n k^2 = 4\cdot\frac{n(n+1)(2n+1)}{6} = \frac{2n(n+1)(2n+1)}{3}

(2)

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

=n(n+1)(2n+7)6= \frac{n(n+1)(2n+7)}{6}

(3)

∑k=1nk(k+1)(k+2)=∑k=1nk3+3k2+2k=∑k=1nk3+3∑k=1nk2+2∑k=1nk\sum_{k=1}^n k(k+1)(k+2) = \sum_{k=1}^n k^3+3k^2+2k = \sum_{k=1}^n k^3 + 3\sum_{k=1}^n k^2 + 2\sum_{k=1}^n k

=n2(n+1)24+3⋅n(n+1)(2n+1)6+2⋅n(n+1)2= \frac{n^2(n+1)^2}{4} + 3\cdot\frac{n(n+1)(2n+1)}{6} + 2\cdot\frac{n(n+1)}{2}

=14n(n+1){n(n+1)+2(2n+1)+4}=14n(n+1)(n2+5n+6)= \frac{1}{4}n(n+1)\{n(n+1)+2(2n+1)+4\} = \frac{1}{4}n(n+1)(n^2+5n+6)

=n(n+1)(n+2)(n+3)4= \frac{n(n+1)(n+2)(n+3)}{4}

(4)

∑k=1n1k(k+2)=∑k=1n12(1k−1k+2)=12(1−13+12−14+13−15+⋯+1n−1n+2)\sum_{k=1}^n \frac{1}{k(k+2)} = \sum_{k=1}^n \frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+2}\right) = \frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\cdots+\frac{1}{n}-\frac{1}{n+2}\right)

=12(1+12−1n+1−1n+2)=12(2(n+1)(n+2)+(n+1)(n+2)−2(n+2)−2(n+1)2(n+1)(n+2))= \frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right) = \frac{1}{2}\left(\frac{2(n+1)(n+2)+(n+1)(n+2)-2(n+2)-2(n+1)}{2(n+1)(n+2)}\right)

=n(3n+5)4(n+1)(n+2)= \frac{n(3n+5)}{4(n+1)(n+2)}

解説: r31bn1z