prove-that-for-n-p-N-2-k-0-p-k-C-n-p-k-C-n-k-n-C-2n-1-p-1-conclude-the-value-of-k-0-n-k-C-n-k-2-

Question Number 73042 by mathmax by abdo last updated on 05/Nov/19

p r o v e t h a t f o r (n, p) \in N^{★^{2}} \sum_{k = 0}^{p} k C_{n}^{p - k} C_{n}^{k} = n C_{2 n - 1}^{p - 1}

c o n c l u d e t h e v a l u e o f \sum_{k = 0}^{n} k {(C_{n}^{k})}^{2}

Answered by mind is power last updated on 05/Nov/19

let p (x) = {(x + 1)}^{n} = \underset{k = 0}{\sum^{n}} C_{n}^{k} X^{k}

p (x) . p^{'} (x) = \underset{k = 0}{\sum^{n}} \underset{j = 1}{\sum^{n}} C_{k}^{n} C_{j}^{n} x^{k} . {jx}^{j - 1}

coeficient of power p - 1 is when k + j = p

= \underset{k = 0}{\sum^{n}} {jC}_{n}^{p - j} C_{n}^{j} X^{p - 1}

p^{2} = {(x + 1)}^{2 n}

Prime causes double exponent: use braces to clarify

coeficent of x^{p - 1} is {nC}_{2 n - 1}^{p - 1}

\Rightarrow \underset{k = 0}{\sum^{p}} {kC}_{n}^{p - k} C_{n}^{k} = {nC}_{2 n - 1}^{p - 1}

if p = n

\Rightarrow Σ k {(C_{n}^{k})}^{2} = {nC}_{2 n - 1}^{n - 1}

Commented by mr W last updated on 06/Nov/19

i l e a r n t a n e w m e t h o d . t h a n k s s i r!

Commented by mind is power last updated on 06/Nov/19

y^{'} re welcom