Question

Sn=1/(1^2+2)+1/(2^2+4)+1/(3^2+6)+...+1/(n^2+2n),求和

Answer 1

因为1/(n^2+2n)=1/n(n+2)=1/2*{1/n-1/(n+2)}

所以Sn=1/2{1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+.+1/n-1/(n+2)}

=1/2{1+1/2-1/(n+1)-1/(n+2)}

=3/4-(2n+3)/2(n+1)(n+2)