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Question Number 151097 by naka3546 last updated on 18/Aug/21

F i n d s u m o f t h i s e x p r e s s i o n .

(\begin{matrix} n \\ 1 \end{matrix}) + 2 (\begin{matrix} n \\ 2 \end{matrix}) + 3 (\begin{matrix} n \\ 3 \end{matrix}) + 4 (\begin{matrix} n \\ 4 \end{matrix}) + \dots + n (\begin{matrix} n \\ n \end{matrix})

P l e a s e s h o w y o u r w o r k i n g s . T h a n k y o u .

Answered by Olaf_Thorendsen last updated on 18/Aug/21

S_{n} = \underset{k = 1}{\sum^{n}} k C_{k}^{n}

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

f^{'} (x) = n {(x + 1)}^{n - 1} = \underset{k = 1}{\sum^{n}} k C_{k}^{n} x^{k - 1}

f^{'} (1) = n 2^{n - 1} = \underset{k = 1}{\sum^{n}} k C_{k}^{n} = S_{n}

Answered by mr W last updated on 18/Aug/21

S_{n} = (\begin{matrix} n \\ 1 \end{matrix}) + 2 (\begin{matrix} n \\ 2 \end{matrix}) + 3 (\begin{matrix} n \\ 3 \end{matrix}) + 4 (\begin{matrix} n \\ 4 \end{matrix}) + \dots + n (\begin{matrix} n \\ n \end{matrix})

S_{n} = 0 (\begin{matrix} n \\ 0 \end{matrix}) + (\begin{matrix} n \\ 1 \end{matrix}) + 2 (\begin{matrix} n \\ 2 \end{matrix}) + 3 (\begin{matrix} n \\ 3 \end{matrix}) + 4 (\begin{matrix} n \\ 4 \end{matrix}) + \dots + n (\begin{matrix} n \\ n \end{matrix})

S_{n} = n (\begin{matrix} n \\ n \end{matrix}) + (n - 1) (\begin{matrix} n \\ n - 1 \end{matrix}) + (n - 2) (\begin{matrix} n \\ n - 2 \end{matrix}) + (n - 3) (\begin{matrix} n \\ n - 3 \end{matrix}) + \dots + 0 (\begin{matrix} n \\ n - n \end{matrix})

S_{n} = n (\begin{matrix} n \\ 0 \end{matrix}) + (n - 1) (\begin{matrix} n \\ 1 \end{matrix}) + (n - 2) (\begin{matrix} n \\ 2 \end{matrix}) + (n - 3) (\begin{matrix} n \\ 3 \end{matrix}) + \dots + 0 (\begin{matrix} n \\ n \end{matrix})

2 S_{n} = n [(\begin{matrix} n \\ 0 \end{matrix}) + (\begin{matrix} n \\ 1 \end{matrix}) + (\begin{matrix} n \\ 2 \end{matrix}) + (\begin{matrix} n \\ 3 \end{matrix}) + \dots + (\begin{matrix} n \\ n \end{matrix})]

2 S_{n} = n \times 2^{n}

\Rightarrow S_{n} = n \times 2^{n - 1}

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