裂项求和即可
4(n+1)/[(n^2+2n)^2]
=[(n+2)²-n²]/[(n+2)²*n²]
=(n+2)²[(n+2)²*n²]-n²/[(n+2)²*n²]
=1/n²-1/(n+2)²
∴ (n+1)/[(n^2+2n)^2]=(1/4)*[1/n²-1/(n+2)²]
∴ Sn=(1/4)[1-1/3²+1/2²-1/4²+1/3²-1/5²+.+1/(n-1)²-1/(n+1)²+1/n²-1/(n+2)²]
=(1/4)[1+1/2²-1/(n+1)²-1/(n+2)²]
裂项求和即可
4(n+1)/[(n^2+2n)^2]
=[(n+2)²-n²]/[(n+2)²*n²]
=(n+2)²[(n+2)²*n²]-n²/[(n+2)²*n²]
=1/n²-1/(n+2)²
∴ (n+1)/[(n^2+2n)^2]=(1/4)*[1/n²-1/(n+2)²]
∴ Sn=(1/4)[1-1/3²+1/2²-1/4²+1/3²-1/5²+.+1/(n-1)²-1/(n+1)²+1/n²-1/(n+2)²]
=(1/4)[1+1/2²-1/(n+1)²-1/(n+2)²]