Prove that for $n\ge 2$, $2{n \choose 2}+{n \choose 1} = n^2$ Problem : Prove that for $n\ge 2$, $2{n \choose 2}+{n \choose 1} = n^2$
My Approach : I would assume that we can prove by induction.
Base case $n=2$.
$$=2{2 \choose 2}+ {2 \choose 1}= (2\cdot 1)+2 =4$$
$$n^2 =2^2 = 4.$$
Assume for $n\ge 2$, $n\ge 2$, $2{n \choose 2}+{n \choose 1} = n^2$  for $n \le k$.
Let $n=k+1$
$$2{k+1 \choose 2}+{k+1 \choose 1} = (k+1)^2$$
And, that's as far as I got.
I get stuck with the ${k+1 \choose 2}$ part.

It's much simpler to give a direct proof:
$$
2 \binom{n}{2} + \binom{n}{1}
=
2 \cdot \frac{n(n-1)}{2} + n
=
n^{2} - n + n
=
n^{2}.
$$
