Computing $\sum_{i=1}^{n}i(i+1)(i+2)$

Question

one method is to expand $n(n+1)(n+2)$ and sum over that, but the answer I note is

$$\sum_{i=1}^{n}i(i+1)(i+2)=\frac{1}{4}n(n+1)(n+2)(n+3)$$

points to an alternate, elegant solution.

Answer 1

First, \begin{align*}\sum_{k=1}^{n}k(k+1)(k+2)&=6\sum_{k=1}^{n}{k+2\choose 3}=6\sum_{k=3}^{n+2}{k\choose 3}=6\sum_{k=1}^{n+2}{k\choose 3}\end{align*} and using the following theorem, we get $$\sum_{k=1}^{n}k(k+1)(k+2)=6{n+3\choose 4}=6\frac{n(n+1)(n+2)(n+3)}{4!}$$

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The above method suggests that $$\sum_{k=1}^{n}k(k+1){\dots}(k+m)=\frac{1}{m+2}n(n+1){\dots}(n+m+1)$$

Answer 2

In general: $$\sum_{k=r}^n\binom{k}{r}=\binom{n+1}{r+1}\tag1$$ This can easily be proved with induction on $n$ using the well known equality: $$\binom{n+1}r+\binom{n+1}{r+1}=\binom{n+2}{r+1}$$

Applying $(1)$ for $r=3$ we find: $$\frac1{3!}\sum_{k=3}^n(k-2)(k-1)k=\frac1{4!}(n-2)(n-1)n(n+1)$$

Taking $n+2$ instead of $n$ and $i$ instead of $k$ gives:$$\sum_{i=1}^ni(i+1)(i+2)=\frac1{4}n(n+1)(n+2)(n+3)$$