Tell me more ×
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.
up vote 3 down vote favorite
1

$$\large \sum_{i = 2}^{25}P(i,2)$$ $P$ 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 Sep 16 '12 at 6:30

3 Answers

up vote 4 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 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
up vote 2 down vote

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

share|improve this answer
up vote 1 down vote

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

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

Your Answer

 
discard

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.