解数学题[(2^3-1)(3^3-1)...100^3-1)]/[(2^3+1)(3^3+1)...(100^3+1)](

上式=[(2-1)(2^2+1^2+1*2)(3-1)(3^2+1^2+1*3)...(100-1)(100^2+1^2+1*100)]/[(2+1)(2^2+1^2-1*2)(3+1)(3^2+1^2-1*3)...(100+1)(100^2+1^2-1*100)]
把各个括号里的数算出来最后消得
=[1*2*(100^2+1^2+1*100)]/3*100*101=3367/