7
$\begingroup$

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 $30$.

$\endgroup$
  • 2
    $\begingroup$ 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... $\endgroup$ – Eric Towers Jun 10 '14 at 4:29
  • $\begingroup$ What is the shortcut? $\endgroup$ – User Not Found Jun 10 '14 at 4:31
  • 3
    $\begingroup$ I see three answers below... $\endgroup$ – Eric Towers Jun 10 '14 at 4:42
11
$\begingroup$

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 $21$

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

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

$\endgroup$
3
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.