Using induction to prove for $n ≥ 1, $ $1 \times 5+2\times6+3\times7 +\cdots +n(n + 4) = \frac 16n(n+1)(2n+13).$ This is a very interesting problem that I came across in an old textbook of mine. So I know its got something to do with mathematical induction, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

Prove that for $n ≥ 1,$
$1 \times 5+2\times6+3\times7 +\cdots +n(n + 4) = \frac 16n(n+1)(2n+13).$


Base Case: For $n = 1$, we have:
$$
1 \times 5 = 5 = \frac{1}{6}(1)(1 + 1)(2(1) + 13)
$$
which works.
Inductive Hypothesis: Assume that the claim holds for $n' = n - 1$, where $n \geq 2$.
It remains to show that the claim holds for $n' = n$. Indeed, observe that:
\begin{align*}
&1 \times 5 +\cdots + n(n + 4) & \\
&= [1 \times 5 + \cdots + (n - 1)(n + 3)] + n(n + 4) &\text{since } n \geq 2 \\
&= \frac{1}{6}(n - 1)(n)(2n + 11) + n(n + 4) &\text{by the inductive hypothesis} \\
&= \frac{1}{6}n [(n - 1)(2n + 11) + 6(n + 4)] \\
&= \frac{1}{6}n [(2n^2 + 9n - 11) + (6n + 24)] \\
&= \frac{1}{6}n [2n^2 + 15n + 13] \\
&= \frac{1}{6}n (n + 1)(2n + 13) \\
\end{align*}
as desired. $~~\blacksquare$
