-1), then when n=k+ 1, there is left =12+32+...+(2k-1) 2+(2k+1) 2 = (.
- 1)+(2k+ 1)^2= 1/3*(2k+ 1)(2k^2-k)+(2k+ 1)^2= 1/3*(2k+ 1)(2k^2-k+6k+3)= 1/3*(2k+ 1) (k+1) (2k+3) =1/3 * (k+1) [4 (k+1) 2-1] = Right. Therefore, for any positive integer n, there is12+3 2+...+(2n-.
-1) hold.