1-Find-the-sum-s-n-1-2x-3x-2-4x-3-nx-n-1-Hence-or-otherwise-find-the-sum-k-1-n-k-2-k-2-Simplify-the-following-i-r-0-n-2r-1-2n-ii-r-0-n-1-r-r-r-n-i

Question Number 37991 by Fawomath last updated on 20/Jun/18

1 . F i n d t h e s u m

s_{n} = 1 + 2 x + 3 x^{2} + 4 x^{3} + \dots + {n x}^{n - 1}

H e n c e, o r o t h e r w i s e, f i n d t h e s u m

\underset{k = 1}{\sum^{n}} k . 2^{k}

2 . S i m p l i f y t h e f o l l o w i n g

i . \underset{r = 0}{\sum^{n}} (_{2 r - 1}^{2 n})

i i . \underset{r = 0}{\sum^{n}} {(- 1)}^{r} r (_{r}^{n})

i i i . \underset{r = 0}{\sum^{n}} {(- 1)}^{r} \frac{1}{r + 1} (_{r}^{n})

i v . \underset{r = 0}{\sum^{n}} (_{2 r}^{2 n})

v . \underset{r = 0}{\sum^{n}} {(- 1)}^{r} (_{n - r}^{n + 1})

3 . F i n d t h e s u m

\underset{r = 0}{\sum^{n - k}} (_{k}^{n - r}), w h e r e k = 0, 1, 2, 3, \dots, n

Commented by prof Abdo imad last updated on 21/Jun/18

1) l e t f (x) = 1 + x + x^{2} + \dots + x^{n} = \sum_{k = 0}^{n} x^{k}

\Rightarrow f^{^{'}} (x) = \sum_{k = 1}^{n} k x^{k - 1} b u t i f x \neq 1

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

s o \sum_{k = 1}^{n} {k x}^{k - 1} = \frac{{n x}^{n + 1} - (n + 1) x^{n} + 1}{{(x - 1)}^{2}}

i f x = 1 f^{^{'}} (x) = 1 + 2 + 3 + \dots + n

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

l e t t a k e x = 2 \Rightarrow \sum_{k = 1}^{n} k . 2^{k - 1} = n 2^{n + 1} - (n + 1) 2^{n} + 1

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

= 4 n 2^{n} - (2 n + 2) 2^{n} + 2 = (2 n - 2) 2^{n} + 2

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

Commented by prof Abdo imad last updated on 21/Jun/18

2) i i l e t p (x) = \sum_{k = 0}^{n} C_{n}^{k} x^{k} w e h a v e

p (x) = {(x + 1)}^{n} \Rightarrow \sum_{k = 0}^{n} {(- 1)}^{k} C_{n}^{k} = p (- 1) = 0

Commented by prof Abdo imad last updated on 21/Jun/18

i i i l e t p (x) = \sum_{k = 0}^{n} C_{n}^{k} x^{k} = {(x + 1)}^{n} \Rightarrow

\int p (x) d x = \sum_{k = 0}^{n} \frac{1}{k + 1} C_{n}^{k} x^{k + 1} + c = \frac{1}{n + 1} {(x + 1)}^{n + 1} + c

x = 0 \Rightarrow c = - \frac{1}{n + 1} \Rightarrow

\sum_{k = 0}^{n} \frac{x^{k + 1}}{k + 1} C_{n}^{k} = \frac{{(x + 1)}^{n + 1} - 1}{n + 1} l e t t a k e x = - 1 \Rightarrow

- \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{k + 1} C_{n}^{k} = \frac{{(x + 1)}^{n + 1} - 1}{n + 1} \Rightarrow

\sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{k + 1} C_{n}^{k} = \frac{1 - {(x + 1)}^{n + 1}}{n + 1} .

Commented by prof Abdo imad last updated on 21/Jun/18

\Rightarrow \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{k + 1} C_{n}^{k} = \frac{1}{n + 1} .

Answered by ajfour last updated on 20/Jun/18

s_{n} = 1 + 2 x + 3 x^{2} + 4 x^{3} + \dots . + {n x}^{n - 1}

{x s}_{n} = [x + 2 x^{2} + 3 x^{3} + \dots . + (n - 1) x^{n - 1} + {n x}^{n}

(1 - x) s_{n} = 1 + x + x^{2} + \dots + x^{n - 1} - {n x}^{n}

s_{n} = \frac{1 - x^{n}}{{(1 - x)}^{2}} - \frac{{n x}^{n}}{1 - x} .

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