find interms of n the sum Σ_(k=0) ^n k^2 C


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Question Number 33074 by prof Abdo imad last updated on 10/Apr/18

$f i n d i n t e r m s o f n t h e s u m \sum_{k = 0}^{n} k^{2} C_{n}^{k}$
Commented by abdo imad last updated on 10/Apr/18

$l e t c o n s i d e r t h e p o l y n o m p (x) = \sum_{k = 0}^{n} C_{n}^{k} x^{k}$ $w e k n o w t h a t p (x) = {(x + 1)}^{n} \Rightarrow p^{^{'}} (x) = n {(x + 1)}^{n - 1} \Rightarrow$ $p^{^{″}} (x) = n (n - 1) {(x + 1)}^{n - 2} f r o m a n o t h e r s i d e$ $p^{^{'}} (x) = \sum_{k = 1}^{n} k C_{n}^{k} x^{k - 1} a n d$ $p^{^{″}} (x) = \sum_{k = 2}^{n} k (k - 1) C_{n}^{k} x^{k - 2} = \sum_{k = 2}^{n} k^{2} C_{n}^{k} x^{k - 2} - \sum_{k = 2}^{n} k C_{n}^{k} x^{k - 2}$ $\Rightarrow \sum_{k = 2}^{n} k^{2} C_{n}^{k} x^{k - 2} = p^{^{″}} (x) + \sum_{k = 2}^{n} k C_{n}^{k} x^{k - 2}$ $x = 1 \Rightarrow \sum_{k = 2}^{n} k^{2} C_{n}^{k} = p^{^{″}} (1) + \sum_{k = 2}^{n} k C_{n}^{k}$ $= n (n - 1) 2^{n - 2} + n 2^{n - 1} - 1 \Rightarrow$ $\sum_{k = 1}^{n} k^{2} C_{n}^{k} = n (n - 1) 2^{n - 2} + n 2^{n - 1} .$ $= (n^{2} - n) 2^{n - 2} + 2 n 2^{n - 2} = (n^{2} + n) 2^{n - 2} .$ $\sum_{k = 1}^{n} k^{2} C_{n}^{k} = (n^{2} + n) 2^{n - 2} .$
	Answered by sma3l2996 last updated on 10/Apr/18

	$\underset{k = 0}{\sum^{n}} k^{2}^{n} C_{k} = ? ?$ $k^{2}^{n} C_{k} = k \times k^{n} C_{k} = k \times n^{n - 1} C_{k - 1}$ $s o \underset{k = 0}{\sum^{n}} k^{2}^{n} C_{k} = n \underset{k = 1}{\sum^{n}} k^{n - 1} C_{k - 1}$ $l e t l = k - 1$ $\underset{k = 1}{\sum^{n}} k^{n - 1} C_{k - 1} = \underset{l = 0}{\sum^{n - 1}} (l + 1)^{n - 1} C_{l}$ $= \underset{l = 0}{\sum^{n - 1}} l^{n - 1} C_{l} + \underset{l = 0}{\sum^{n - 1}}^{n - 1} C_{l}$ $w e a l r e a d y k n e w t h a t \underset{k = 0}{\sum^{n}}^{n} C_{k} = 2^{n}$ $a n d \underset{l = 0}{\sum^{n}} l^{n} C_{l} = 2^{n - 1} n (c h e c k Q 33072)$ $S o \underset{k = 0}{\sum^{n}} k^{2}^{n} C_{k} = n (\underset{k = 0}{\sum^{n - 1}} k^{n - 1} C_{k} + \underset{k = 0}{\sum^{n - 1}}^{n - 1} C_{k})$ $= n (2^{n - 2} (n - 1) + 2^{n - 1})$ $\underset{k = 0}{\sum^{n}} k^{2}^{n} C_{k} = 2^{n - 2} n (n + 1)$
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