用数学归纳法证明:1^2+2^2+…+n^2=n(n+1)(2n+1)/6(n是正整数）.

当n=1时,左边=1^2=1右边=1*(1+1)*(2+1)/6=1相符；设n=k时成立即：1^2+2^2+…+k^2=k(k+1)(2k+1)/6则1^2+2^2+…+k^2+(k+1)^2=k(k+1)(2k+1)/6+(k^2+2k+1)=(2k^3+3k^2+k+6k^2+12k+6)/6=(k+1)(k+2)(2k+3)/6=(k+1)[(k+1)+1...