Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\large \sum_{i = 2}^{25}P(i,2)$$ $P$ stands for "permutations".

share|improve this question
So do you know a formula for $P(i,2)$? Do you know formulas for $\sum_{i=1}^n i$ and $\sum_{i=1}^n i^2$? –  Robert Israel Sep 16 '12 at 6:28
Yes,$$ P(i,2) = {i! \over (i - 2)!}$$ Second involves the sum of an arithmetic sequence.$${n(n + 1) \over 2}$$Third:$${n(n + 1)(2n + 1)\over 6} $$ –  Parth Kohli Sep 16 '12 at 6:30
add comment

3 Answers

up vote 5 down vote accepted

$$\displaystyle\sum_{i=2}^{25} P(i,2) = \displaystyle\sum_{i=2}^{25} \frac{i!}{(i-2)!} = \displaystyle\sum_{i=2}^{25} \frac{i (i-1) (i-2)!}{(i-2)!} = \displaystyle\sum_{i=2}^{25} i (i-1) = \displaystyle\sum_{i=2}^{25} (i^2 - i) = \displaystyle\sum_{i=2}^{25} i^2 - \displaystyle\sum_{i=2}^{25} i$$

share|improve this answer
That was very good. Thanks, I can do the rest :) –  Parth Kohli Sep 16 '12 at 6:36
There's a typo in the fourth and proceeding expressions: $\sum_{i=2}^{25}i(i-i)$. –  000 Sep 23 '12 at 22:23
The fourth summation says $$\sum_{i=2}^{25}i(i-i).$$ I take it that you meant $$\sum_{i=2}^{25}i(i-1).$$ Am I misunderstanding? If so, sorry. –  000 Sep 24 '12 at 0:03
@Limitless: You're right. –  Rod Carvalho Sep 24 '12 at 0:06
add comment

$$i(i-1)=\frac{1}{3}\Big((i+1)(i)(i-1)-(i)(i-1)(i-2)\Big).$$ Add up from $i=2$ to $i=25$, and observe the beautiful cancellations (telescoping).

share|improve this answer
add comment


There are well-known formulas for $\sum_{i=1}^ni$ and $\sum_{i=1}^ni^2$ that you can use to finish the job; these formulas can be found (among many other places) in most standard calculus texts when summations are introduced preparatory to doing Riemann sums.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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