# Help to compute sum of products

I need to compute the following sum:

$$(1\times2\times3)+(2\times3\times4)+(3\times4\times5)+ ...+(20\times21\times22)$$

All that I have deduced is:

1. Each term is divisible by $6$. So sum is is divisible by $6$.
2. Sum is divisible by $5$ as 1st term is $1$ less than multiple of $5$ and second term is $1$ more than multiple of $5$. Next three terms are divisible by $5$. This cycle continues for every $5$ terms.

So sum will obviously be divisible by $30$.

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There is a shortcut, ..., but this sum only has 20 terms, all of which you can write out by hand or with a pocket calculator pretty quickly... – Eric Towers Jun 10 '14 at 4:29
What is the shortcut? – Arghya Chakraborty Jun 10 '14 at 4:31
I see three answers below... – Eric Towers Jun 10 '14 at 4:42

HINT:

$$(r-1)r(r+1)=r^3-r$$

Now, $$\sum_{r=1}^n r=\frac{n(n+1)}2$$ and $$\sum_{r=1}^n r^3=\left(\frac{n(n+1)}2\right)^2$$

Here $r=1$ to $21$

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Hint: Note that $(n+1)(n+2)(n+3)(n+4)-(n)(n+1)(n+2)(n+3)=4(n+1)(n+2)(n+3)$.

Using this identity write our sum as a collapsing (telescoping) sum. It may help to look at $4$ times our sum.

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There is a very short solution, using difference calculus, which is a theory underlying Andre Nicolas' hint.

$$\sum_{k=0}^{23}k^{\underline{3}}\delta k=\frac{1}{4}k^{\underline{4}}|_0^{23}=\frac{1}{4}(23^{\underline{4}}-0^{\underline{4}})=\frac{23\cdot 22\cdot 21\cdot 20}{4}=53,130$$

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Here's an interesting solution:

$(1\cdot2\cdot3)+(2\cdot3\cdot4)+(3\cdot4\cdot5)+\dots+(20\cdot21\cdot22)$

$\dfrac{3!}{0!}+\dfrac{4!}{1!}+\dfrac{5!}{2!}+\dots+\dfrac{22!}{19!}$

$3!\left(\dfrac{3!}{0!3!}+\dfrac{4!}{1!3!}+\dfrac{5!}{2!3!}+\dots+\dfrac{22!}{19!3!}\right)$

$3!\left(\dbinom{3}{3}+\dbinom{4}{3}+\dbinom{5}{3}+\dots+\dbinom{22}{3}\right)$

then using the hockey-stick identity, we see that this is equal to

$3!\dbinom{23}{4} = 3!\dfrac{23\cdot22\cdot21\cdot20}{4\cdot3!} = \dfrac{23\cdot22\cdot21\cdot20}{4} = 53130$

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