<s>[INST] Prove by mathematical induction Prove the following statement by using mathematical induction:                               $3+8+ ··· + (n^2 − 1) = \frac{1}{6}n(n − 1)(2n + 5)$, for all integers $n ≥ 2$.
After making a common factor $\frac{1}{6}k$ and expanding I'm left with $2k^2+9k-5 = 2k^2+9k+7$. Where am I going wrong?                                        
 [/INST] 
After making a common factor $\frac{1}{6}k$ and expanding I'm left with $2k^2+9k-5 = 2k^2+9k+7$.

It looks like you lost a term somewhere along the way. Compare to:
$$
\require{cancel}
\begin{align}
3+8+ ··· + (n^2 − 1)+\left((n+1)^2-1\right) &= \frac{1}{6}n(n − 1)(2n + 5) \;+\; n^2+2n + \bcancel{1}-\bcancel{1} \\
 &= \frac{1}{6}n(2n^2+3n-5) + n(n+2) \\
 &= \frac{1}{6}n(2n^2+3n-5+6n+12) \\
 &= \frac{1}{6}n(2n^2+9n+7) \\
 &= \frac{1}{6}n(n+1)(2n+7) \\
\end{align}
$$
</s>