在线等！！数列1/1^2+2,1/2^2+4,1/3^2+6,1/4^2+8……的前n项和是多少

数列第n项为：1/(n^2+2n)=1/n(n+2)=(1/2)(1/n-1/(n+2))
因此数列前n项和为：
(1/2)[(1-1/3)+(1/2-1/4)+(1/3-1/5)+……+(1/(n-2)-1/n)+(1/(n-1)-1/(n+1))+(1/n-1/(n+2))]
=(1/2)[1+1/2-1/(n+1)-1/(n+2)]
=(1/2)[3/2-(2n+3)/(n+1)(n+2)]
=(1/2)[3(n^2+n-1)/2(n+1)(n+2)]
=3(n^2+n-1)/4(n+1)(n+2)
{或=(3n^2+3n-3)/(4n^2+12n+8)}

1/n^2+2n
这是通项式
前n项和自己对照书里面求

1/1^2+2,1/2^2+4,1/3^2+6,1/4^2+8……+1/n(n+2)
=1/2[1-1/2+1/2-1/3+1/3-1/4+------+1/n-1/(n+1)]
=1/2[1-1/(n+1)]
=n/2(n+1)

1/1^2+2=2
2*n=和