∵1^2+2^2+...+n^2=n(n+1)(2n+1)/6
∴3/(1^2)+5/(1^2+2^2)+7/(1^2+2^2+3^2)+...+(2n+1)/(1^2+2^2+...+n^2)
=6×3/(1×2×3)+6×5/(2×3×5)+6×7/(3×4×7)+...+6(2n+1)/[n(n+1)(2n+1)]
=6{1/(1×2)+1/(2×3)+1/(3×4)+...+1/[n(n+1)]}
=6[1-1/2+1/2-1/3+1/3-1/4+...+1/n-1/(n+1)]
=6n/(n+1)