Sn=1/(1^2+2)+1/(2^2+4)+1/(3^2+6)+.+1/(n^2+2n),求Sn的通项公式
问题描述:
Sn=1/(1^2+2)+1/(2^2+4)+1/(3^2+6)+.+1/(n^2+2n),求Sn的通项公式
答
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)