Question

1/(1^2+2),1/(2^2+4),1/(3^2+6),...的前n项和等于?

Answer 1

通项为an=1/(n²+2n)=1/n(n+2)

=1/2[1/n-1/(n+2)]

故前n项和=1/2[1+1/2-1/(n+1)-1/(n+2)]

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