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！
