Mathematical Induction Problem with Fraction

Question

$$(3n-2)^2=\frac{n(6n^2-3n-1)}{2}$$

I can't seem to solve it out to the point where I can prove it right or wrong. I always hit some sort of roadblock where I don't have enough info to prove it wrong, but I can't move farther to prove it right. Help?

Answer 1

As stated, this can't possibly be true for infinitely many $n$. The LHS is a quadratic polynomial but the RHS is a cubic. This means you have a cubic equation, so it can be satisfied by at most three distinct values of $n$.

Answer 2

I think the problem is misstated. You might have better luck proving (by induction) that for all $n\ge 1$, $\sum_{k=1}^n (3k-2)^2 = \frac{n(6n^2-3n-1)}{2}$