Prove by induction: $$\sum_{i=1}^n \binom{i}{2}= {n+1 \choose 3}$$

I already know that: $$\sum_{i=1}^n \binom{i}{2} = {i+1 \choose 2+1}$$

And the LHS is now equal: $$\sum_{i=1}^n \binom{i}{2} + {i+1 \choose 2+1}$$

How do I continue to solve by induction?

  • 2
    $\begingroup$ Something is not right with what you "already know". The variable $i$ is the index of summation, and it has no meaning on the right side of the second equation. $\endgroup$ – hardmath Nov 21 '15 at 4:53

Suppose for a fixed $n$ that $\color{blue}{\sum_{i=1}^n \binom{i}{2}= {n+1 \choose 3}}$.

Consider $\sum_{i=1}^{n+1} \binom{i}{2}$. Then

$$\sum_{i=1}^{n+1} \binom{i}{2}= \binom{n+1}{2} + \color{blue}{\sum_{i=1}^n \binom{i}{2}} = \binom{n+1}{2} + \color{blue}{\binom{n+1}{3}}.$$

Is the right hand side equal to $\binom{(n+1)+1}{3}$? If it is, then we've proved that the claim holds for $n+1$.


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