# Summation of a finite series involving permutations.

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

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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

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

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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

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).

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

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.

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