求数列(1/1^2+2)+(1/2^2+4)+...+(1/n^2+2n)的前n项和
问题描述:
求数列(1/1^2+2)+(1/2^2+4)+...+(1/n^2+2n)的前n项和
答
(1/1^2+2)+(1/2^2+4)+...+(1/n^2+2n)=(1/2)[(1/1)-(1/3)]+(1/2)[(1/2)-(1/4)]+(1/2)[(1/3)-(1/5)]+......+(1/2)[1/(n-1)-1/(n+1)]+(1/2)[1/n-1/(n+2)]=(1/2)[1+1/2-1/(n+1)-1/(n+2)]=(1/2)[3/2-(2n+3)/(n+1)...