자연계열 1-3번

[1.3] 자연수 $\displaystyle k$에 대하여$\displaystyle \int_{-1}^1x^{2k-1}\,dx=\Biggl[\frac1{2k}x^{2k}\Biggr]_{-1}^1=\frac1{2k}\{1^{2k}-(-1)^{2k}\}=0$이므로$\displaystyle \begin{aligned}a_{2n}&=\int_{-1}^1\{2+6x+12x^2+20x^3+\cdots+(2n-1)\times2nx^{2n-2}+2n\times(2n+1)x^{2n-1}\}\,dx\\&=\int_{-1}^1\{2+12x^2+\cdots+(2n-1)\times2nx^{2n-2}\}\,dx\\&=\Biggl[2x+4x^3+\cdots+2nx^{2n-1}\Biggr]_{-1}^1=(2+4+\cdots+2n)-(-2-4-\cdots-2n)=2n(n+1)\end{aligned}$이다.

$\displaystyle b_n=\frac{a_{2n}}n=2(n+1)$이므로

$\displaystyle b_1+b_2+b_3+\cdots+b_{25}=\sum_{n=1}^{25}2(n+1)=2\sum_{n=1}^{25}n+\sum_{n=1}^{25}2=2\times\frac{25\times26}{2}+2\times25=700$이다.