第1章 1.10

(1)

$\sum_{k=1}^{5}(2k-3) = -1+1+3+5+7 = 2\sum_{k=1}^{5}k - \sum_{k=1}^{5}3 = 2\cdot\frac{5(5+1)}{2} - 3\cdot 5 = 30-15 = 15$

(2)

$\sum_{k=0}^{5}2^{k} = 1+2+2^{2}+2^{3}+2^{4}+2^{5} = 1+\frac{2(1-2^{5})}{1-2} = 1+\frac{2(1-32)}{-1} = 1+62 = 63$

(3)

$\sum_{k=1}^{n}k(k+1) = 1\cdot 2+2\cdot 3+3\cdot 4+\cdots n(n+1) = \sum_{k=1}^{n}k^{2}+\sum_{k=1}^{n}k$

$= \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} = \frac{n(n+1)\{(2n+1)+3\}}{6} = \frac{n(n+1)(2n+4)}{6} = \frac{n(n+1)(n+2)}{3}$

(4)

$\sum_{k=1}^{n}(k^{3}+2) = (1^{3}+2)+(2^{3}+2)+(3^{3}+2)+\cdots+(n^{3}+2) = \sum_{k=1}^{n}k^{3}+\sum_{k=1}^{n}2 = \frac{n^{2}(n+1)^{2}}{4}+2n$