原题是:{1/(n²+n)}的前n项和怎么求?
1/(n²+n)=(1/n)-(1/(n+1))
1/(1²+1)+1/(2²+2)+...+1/(n²+n)
=(1-(1/2))+((1/2)-(1/3))+...+((1/n)-(1/(n+1)))
=1-(1/(n+1))
=n/(n+1)
所以 {1/(n²+n)}的前n项和是:n/(n+1)
希望能帮到你!