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?
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:
$$ \require{cancel} \begin{align} 3+8+ ··· + (n^2 − 1)+\left((n+1)^2-1\right) &= \frac{1}{6}n(n − 1)(2n + 5) \;+\; n^2+2n + \bcancel{1}-\bcancel{1} \\ &= \frac{1}{6}n(2n^2+3n-5) + n(n+2) \\ &= \frac{1}{6}n(2n^2+3n-5+6n+12) \\ &= \frac{1}{6}n(2n^2+9n+7) \\ &= \frac{1}{6}n(n+1)(2n+7) \\ \end{align} $$
