How to prove: If $n \in N$, then $n^2=2\binom{n}{2} + \binom{n}{1}$

Question

Prove using a direct proof method: If $n \in N$, then $n^2=2\binom{n}{2} + \binom{n}{1}$.

I would assume that the proof uses the Binomial Theorem, however I don't know how to apply it in this context.

Perhaps it is something in regard to:

$$n^2 = (2n-n)^2$$ $$n^2 = \binom{n}{0}(2n)^n(-n)^0 + \binom{n}{1}(2n)^{n-1}(-n) + \binom{n}{2}(2n)^{n-2}n^2$$ $$n^2 = (2n)^n - n^2(2n)^n \frac{1}{(2n)} + \frac{n^3(n+1)}{2}(2n)^n \frac{1}{(2n)^2} $$ $$...$$

Answer 1

Use the fact that $2\binom n2=2\frac{n(n-1)}2=n^2-n.$

Answer 2

There is a typo in your working: It should be:

$$n^2 = (2n-n)^2$$ $$n^2 = \binom{n}{0}(2n)^2(-n)^0 + \binom{n}{1}(2n)(-n) + \binom{n}{2}(2n)^{0}n^2$$