Prove that $\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+...+\binom{n}{2}=\binom{n+1}{3}$

Question

I don't know how to begin. I need to prove it for $n\geq 2$. I'm not quite good at this. It's my weak point. Thank you for any suggestions.

$$\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+...+\binom{n}{2}=\binom{n+1}{3}$$

Answer 1

You can also use induction.

For $n=2$, it is true since $\binom{2}{2}=\binom{2+1}{3}$.

If for $n=j$ the equation is true, then it is also true for $n=j+1$ since

$\binom{2}{2}+\binom{3}{2}+ ...+\binom{j}{2}+\binom{j+1}{2}=\binom{j+1}{3}+\binom{j+1}{2}=\binom{j+1+1}{3}$

Therefore, it is true for all $n\geq 2$

Answer 2

Coefficient of $x^2$ in $\displaystyle (1+x)^2+(1+x)^3+\cdots +(1+x)^n=\frac{(1+x)^2-(1+x)^{n+1}}{1-(1+x)}$

Coefficient of $x^3$ in $\displaystyle (1+x)^{n+1}-(1+x)^2=\binom{n+1}{3}$