Question

化简 ,下面的数都在二次根号内,(n为正整数),1/n^2+1/(n+1)^2+1/(2n+1)^2,

Answer 1

设a=n,b=n+1,有b-a=1

则通分得分子为：

(ab)^2+a^2(a+b)^2+b^2(a+b)^2

=a^2b^2+(a+b)^2(a^2+b^2)

=a^2b^2+[(a-b)^2+4ab][(a-b)^2+2ab]

=a^2b^2+(1+4ab)(1+2ab)

=a^2b^2+1+6ab+8a^2b^2

=9a^2b^2+6ab+1

=(3ab+1)^2

所以原根式=(3ab+1)/[ab(a+b)]=[3n(n+1)+1]/[n(n+1)/(2n+1)]