prove-that-n-N-k-0-n-k-C-2n-n-k-nC-2n-1-n-

Question Number 73046 by mathmax by abdo last updated on 05/Nov/19

p r o v e t h a t \forall n \in N \sum_{k = 0}^{n} k C_{2 n}^{n + k} = {n C}_{2 n - 1}^{n}

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

\underset{k = 0}{\sum^{n}} C_{2 n}^{k + n} = \underset{k = 0}{\sum^{n}} C_{2 n}^{n - k} = \underset{k = 0}{\sum^{n}} C_{2 n}^{k}

\underset{k = 0}{\sum^{k = n}} C_{2 n}^{k + n} = \underset{k = n}{\sum^{2 n}} C_{2 n}^{k}

\Rightarrow \underset{k = 0}{\sum^{n}} C_{2 n}^{k} + \underset{k = n}{\sum^{2 n}} C_{2 n}^{k} = \underset{k = 0}{\sum^{2 n}} C_{2 n}^{k} + C_{2 n}^{n} = 2^{2 n} + C_{2 n}^{n}

\Rightarrow \underset{k = 0}{\sum^{n}} C_{2 n}^{k + n} = 2^{2 n - 1} + \frac{C_{2 n}^{n}}{2}

\underset{k = 0}{\sum^{n}} {kC}_{2 n}^{k + n} = \underset{k = 0}{\sum^{n}} (k + n - n) C_{2 n}^{n + k} = \underset{k = 0}{\sum^{n}} (k + n) C_{2 n}^{n + k} - n \underset{k = 0}{\sum^{n}} C_{2 n}^{k + n}

Σ (k + n) C_{2 n}^{n + k} = \underset{k = 0}{\sum^{n}} (k + n) . \frac{2 n!}{(k + n)! . (n - k)!}

= \underset{k = 0}{\sum^{n}} . \frac{2 n . (2 n - 1)!}{(n + k - 1)! . (2 n - 1 - (n + k - 1)!} = 2 n \underset{k = 0}{\sum^{n}} C_{2 n - 1}^{n + k - 1}

\underset{k = 0}{\sum^{n}} C_{2 n - 1}^{n + k - 1} = \underset{k = 0}{\sum^{n}} C_{2 n - 1}^{n - k}

\underset{k = 0}{\sum^{n}} C_{2 n - 1}^{n + k - 1} + \underset{k = 0}{\sum^{n}} C_{2 n - 1}^{n - k} = \underset{j = 0}{\sum^{2 n - 1}} C_{2 n - 1}^{j} + C_{2 n - 1}^{n} + C_{2 n - 1}^{n - 1}

\Rightarrow \underset{k = 0}{\sum^{n}} C_{2 n - 1}^{n + k - 1} = \frac{1}{2} (2^{2 n - 1} + (C_{2 n - 1}^{n} + C_{2 n - 1}^{n - 1})

= 2^{2 n - 2} + \frac{C_{2 n}^{n}}{2}

Σ (k + n) C_{2 n - 1}^{n + k} = 2 n . (2^{n - 2} + \frac{C_{2 n}^{n}}{2})

\Rightarrow \underset{k = 0}{\sum^{n}} {kC}_{2 n}^{n + k} = \underset{k = 0}{\sum^{n}} (k + n) C_{2 n}^{n + k} - n \underset{k = 0}{\sum^{n}} C_{2 n}^{n + k}

= 2 n (2^{n - 2} + \frac{C_{2 n}^{n}}{2}) - n . (2^{n - 1} + \frac{C_{2 n}^{n}}{2})

= \frac{n}{2} . C_{2 n}^{n} = \frac{n}{2} C_{2 n}^{n} = \frac{n}{2} . \frac{2 n!}{n! . n!} = \frac{n . 2 n . (2 n - 1)!}{2 . n . (n - 1)! . n!} = \frac{n . (2 n - 1)!}{n! . (2 n - 1 - n)!}

= {nC}_{2 n - 1}^{n}

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