用拆项法求数列(2^2+1)/(2^2-1),(3^2+1)/(3^2-1),(4^2+1)/(4^2-1),…的前n项之和.
问题描述：
用拆项法求数列(2^2+1)/(2^2-1),(3^2+1)/(3^2-1),(4^2+1)/(4^2-1),…的前n项之和.
答
a(n) = [(n+1)^2 + 1]/[(n+1)^2 - 1] = 1 + 2/[(n+1)^2 - 1]=1 + 2/[(n+2)n]=1 + 1/n - 1/(n+2),s(n) = a(1)+a(2)+a(3)+a(4)+...+a(n-3)+a(n-2)+a(n-1)+a(n)=n + [1/1-1/3 + 1/2-1/4 + 1/3-1/5 + 1/4-1/6 + ...+ 1/(...