Here are my problems: For $n \geq 2$ and $n$ is integer. Prove that:$$ C_2^2+C_4^2+ C_6^2 + \cdots + C_{2n}^2 = \frac{n(n+1)(4n-1)}{6}$$
without using induction.
I have tried to calculate: $$C_2^1+C_2^2+C_3^2+\cdots+C_{2n}^2$$
But it seems to be wrong.
By binomial theorem,
$$\frac{(1+x)^{2n}+ (1-x)^{2n}}{2}= \sum_{r=0}^{n} \binom {2n}{2r}x^{2r}$$
Squaring both side, we have :
$$\left(\frac{(1+x)^{2n}+ (1-x)^{2n}}{2}\right)^2= \left(\sum_{r=0}^{n} \binom {2n}{2r}x^{2r}\right)\left(\sum_{r=0}^{n} \binom {2n}{2r}x^{2n-2r}\right)$$
Can you see that how the coefficient of an appropriate power of $x$ in above product is related to your sum ?
