n-0-3-n-1-5-n-2-2n-1-n-n-n-1-2-n-2-3-n-3-n-n-n-23-11-n-

Question Number 208662 by efronzo1 last updated on 20/Jun/24

\frac{(\begin{matrix} n \\ 0 \end{matrix}) + 3 (\begin{matrix} n \\ 1 \end{matrix}) + 5 (\begin{matrix} n \\ 2 \end{matrix}) + \dots + (2 n + 1) (\begin{matrix} n \\ n \end{matrix})}{(\begin{matrix} n \\ 1 \end{matrix}) + 2 (\begin{matrix} n \\ 2 \end{matrix}) + 3 (\begin{matrix} n \\ 3 \end{matrix}) + \dots + n (\begin{matrix} n \\ n \end{matrix})} = \frac{23}{11}

n = ?

Answered by Berbere last updated on 20/Jun/24

A = \underset{k = 0}{\sum^{n}} (2 k + 1) (\begin{matrix} n \\ k \end{matrix}); \underset{k = 0}{\sum^{n}} x^{2 k + 1} (\begin{matrix} n \\ k \end{matrix}) = f (x)

= x {(1 + x^{2})}^{n} \Rightarrow A = f^{'} (1)

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

B = \underset{k = 0}{\sum^{n}} k (\begin{matrix} n \\ k \end{matrix}); g (x) = Σ (\begin{matrix} n \\ k \end{matrix}) x^{k} = {(1 + x)}^{n}

B = g^{'} (1) = n . 2^{n - 1}

\Leftrightarrow \frac{(n + 1) 2^{n}}{n . 2^{n - 1}} = \frac{23}{11} \Leftrightarrow \frac{2 (n + 1)}{n} = \frac{23}{11} \Rightarrow n = 22

Answered by Berbere last updated on 20/Jun/24

\underset{k = 1}{\sum^{n}} k (\begin{matrix} n \\ k \end{matrix}) = \frac{k . n!}{k! . (n - k)!} = n \sum_{k ⩾ 1} \frac{(n - 1)!}{(k - 1)! . (n - 1 - (k - 1))!}

= n \underset{k = 0}{\sum^{n - 1}} (\begin{matrix} n \\ k \end{matrix}) = n 2^{n - 1} . . S

\underset{k = 0}{\sum^{n}} (a k + b) (\begin{matrix} n \\ k \end{matrix}) = a S + b . 2^{n}

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