1²+2²+3²+...+n²=n(n+1)(2n+1)/6
1³+2³+3³+...+n³=n²(n+1)²/4
所以(1²+2²+3²+...+n²)/(1³+2³+3³+...+n³)=2(2n+1)/[3n(n+1)]
(1²+2²+3²+...+n²)/(1³+2³+3³+...+n³)-[1²+2²+3²+...+(n+1)²]/[1³+2³+3³+...+(n+1)³]=2(2n+1)/[3n(n+1)]-2(2n+3)/[3(n+1)(n+2)]=4/[3n(n+2)]=(2/3)*[1/n-1/(n+2)].
原式=(2/3)*[(1-1/3)+(1/2-1/4)+(1/3-1/5)+...+(1/24-1/26)]
=(2/3)*(1+1/2-1/25-1/26)
=308/325.
