Prove by induction: $C(2n, 2) = 2C(n, 2) + n^2$ 
Show that if $n$ is a positive integer, then $C(2n, 2) = 2C(n, 2) + n^2$. Here, $C(a, b)$ means the binomial coefficient $\dbinom{a}{b}$.
Prove this by induction.

Here is my calculation:
$n$ cannot be $1$ because $n$ should be equal or larger than $2$ for $C(n, 2)$.
If $n=2$: $C(4, 2) = 2C(2, 2) + 2^2 = 6$
If $n=m$: $C(2m, 2) = 2C(m, 2) + m^2 = 2m^2 - m$
If $n=m+1$: $C(2m+2, 2) = 2C(m+1, 2) + (m+1)^2 = 2m^2 + 3m + 1$
And I think I have to prove that:
\begin{align*}
C(2m+2, 2) & = [\dots]\\
           & = [2C(m, 2) + m^2] + [2C(m, 1) + (2m + 1)]\\
           & = 2C(m+1, 2) + (m+1)^2 && \text{(by Pascal's Identity).}
\end{align*}
But I have no idea what to do on $[\dots]$ part.
Thanks.

Hint:
Show and then use that
$$\binom nk=\binom{n-1}{k-1}+\binom{n-1}k$$
