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How to prove the existence of constants A and B by mathematical induction, so that12+3 2+... ...
When n= 1 = 1/3× (a+b) When n = 2, it is 1+9= 1/3×2(4a+b). If it holds, it is a = 4 and b =-. left = 1+9+25 = 35 = 1/3×3×(4×9- 1)= right holds。 Suppose there is 12+32+ when n = k ≥ 2. ...

-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.