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Question Number 178500 by infinityaction last updated on 17/Oct/22

Commented by abdullahoudou last updated on 17/Oct/22

$i t w o u l d b e n i c e t o g i v e t h e r e s t o f q u e s t i o n s i n s t e a d o f o n l y n u m b e r 15 . . .$
	Answered by mindispower last updated on 17/Oct/22

	$== \sum_{k ⩾ 0} \frac{k (k - 1) n!}{(n - k)! k!}$ $= \underset{k = 0}{\sum^{n}} k (k - 1) C_{k}^{n}$ $\underset{k = 0}{\sum^{n}} x^{k} C_{n}^{k} = {(1 + x)}^{n}$ $\Rightarrow \sum_{k ⩾ 1} {k x}^{k - 1} C_{k}^{n} = n {(1 + x)}^{n - 1}$ $Σ k (k - 1) x^{k - 2} = n (n - 1) {(1 + x)}^{n - 2}$ $Σ k (k - 1) C_{n}^{k} = Σ \frac{n!}{(n - k)! (k - 2)!} = n (n - 1) 2^{n - 2}$
	Commented by infinityaction last updated on 17/Oct/22

	$t h a n k y o u s i r$
	Commented by mindispower last updated on 02/Nov/22

	$w i t h e p l e a s u r$
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