Alternative proof by Induction

Question

I have to prove by induction, for $n\ge 1$ , this equality:

\begin{gather}\sum_{i=1}^{2n} \frac{i(i+1)}{8}=\sum_{i=1}^{n} i^{2} \end{gather}

For it, i use \begin{gather}\sum_{i=1}^{n} i^{2}=\frac{n(n+1)(2n+1)}{6} \end{gather}

My question is: there is a way to prove the first equality by Induction , without the use of the second equality, as i have done? Thanks.

Answer 1

Hint:$$\frac{(2n-1)2n}8+\frac{2n(2n+1)}8=\frac n4\bigl((2n-1)+(2n+1)\bigr)=n^2.$$