if-1-1-x-n-m-0-a-m-x-m-determinate-a-m-

Question Number 98940 by mathmax by abdo last updated on 17/Jun/20

if \frac{1}{{(1 + x)}^{n}} = \sum_{m = 0}^{\infty} a_{m} x^{m} determinate a_{m}

Commented by mr W last updated on 17/Jun/20

f o r n ⩾ 2 :

\frac{1}{{(1 + x)}^{n}} = \underset{m = 0}{\sum^{\infty}} {(- 1)}^{m} C_{n - 1}^{m + n - 1} x^{m}

Answered by maths mind last updated on 17/Jun/20

\frac{\partial^{n}}{\partial x^{n}} \frac{1}{(1 + x)} = \frac{{(- 1)}^{n} . n!}{{(1 + x)}^{n + 1}}

\Rightarrow \frac{1}{{(1 + x)}^{n}} = . \frac{{(- 1)}^{n}}{n!} \frac{\partial^{n - 1}}{\partial x^{n - 1}} . \frac{1}{(1 + x)}

= \frac{{(- 1)}^{n}}{n!} . \frac{\partial^{n - 1}}{\partial x^{n - 1}} . \frac{1}{1 + x} = \frac{{(- 1)}^{n}}{n!} . \frac{\partial^{n - 1}}{\partial x^{n - 1}} (\sum_{p ⩾ 0} {(- 1)}^{p} x^{p})

n ⩾ 2

\frac{1}{{(1 + x)}^{n}} = \frac{{(- 1)}^{n}}{n!} \sum_{p ⩾ n - 1} {(- 1)}^{p} \underset{k = 0}{\prod^{n - 2}} . (p - k) x^{p - (n - 1)}

n = 0

\frac{1}{{(1 + x)}^{0}} = 1

n = 1

\frac{1}{(1 + x)} = \sum_{n ⩾ 0} {(- 1)}^{n} x^{n}

Commented by mathmax by abdo last updated on 18/Jun/20

thank you sir .

Answered by mathmax by abdo last updated on 18/Jun/20

f (x) = \frac{1}{{(1 + x)}^{n}} \Rightarrow f (x) = \sum_{m = 0}^{\infty} \frac{f^{(m)} (0)}{m!} x^{m} let determine f^{(m)} (0)

f (x) = {(x + 1)}^{- n} = {(x + 1)}^{p} with p = - n \Rightarrow f^{(m)} (x) = {{(x + 1)}^{p}}^{(m)}

if m > p f^{(m)} (x) = 0 if m ⩽ p we have f^{(1)} (x) = p {(x + 1)}^{p - 1}

f^{(2)} (x) = p (p - 1) {(x + 1)}^{p - 2} \Rightarrow f^{(m)} (x) = p (p - 1) \dots (p - m + 1) {(x + 1)}^{p - m}

= (- n) (- n - 1) \dots . (- n - m + 1) {(x + 1)}^{- n - m}

= {(- 1)}^{m} n (n + 1) \dots . (n + m - 1) {(x + 1)}^{- n - m} \Rightarrow

f (x) = \sum_{m = 0}^{n} \frac{f^{(m)} (0)}{m!} x^{m} = \sum_{m = 0}^{n} \frac{{(- 1)}^{m} n (n + 1) \dots (n + m - 1)}{m!} x^{m} \Rightarrow

a_{m} = \frac{{(- 1)}^{m} n (n + 1) \dots . (n + m - 1)}{m!}