if (1/((1+x)^n )) =Σ_(m=0) ^∞ a_m x^m determinate a


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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
	$f(x) =(1/((1+x)^n )) ⇒f(x) =Σ_(m=0) ^∞ ((f^((m)) (0))/(m!)) x^m let determine f^((m)) (0) f(x) =(x+1)^(−n) =(x+1)^p with p =−n ⇒f^((m)) (x) ={(x+1)^p }^((m)) if m>p f^((m)) (x)=0 if m≤p we have f^((1)) (x) =p(x+1)^(p−1) f^((2)) (x) =p(p−1)(x+1)^(p−2) ⇒f^((m)) (x) =p(p−1)...(p−m+1)(x+1)^(p−m) =(−n)(−n−1)....(−n−m+1)(x+1)^(−n−m) = (−1)^m n(n+1)....(n+m−1)(x+1)^(−n−m) ⇒ f(x) =Σ_(m=0) ^n ((f^((m)) (0))/(m!)) x^m =Σ_(m=0) ^n (((−1)^m n(n+1)...(n+m−1))/(m!)) x^m ⇒ a_m =(((−1)^m n(n+1)....(n+m−1))/(m!))$
	$f (x) = \frac{1}{{(1 + x)}^{n}} \Rightarrow f (x) = \sum_{m = 0}^{\infty} \frac{f^{(m)} (0)}{m!} x^{m} let determine f^{(m)} (0)$ $f (x) = {(x + 1)}^{- n} = {(x + 1)}^{p} with p = - n \Rightarrow f^{(m)} (x) = {{(x + 1)}^{p}}^{(m)}$ $if m > p f^{(m)} (x) = 0 if m ⩽ p we have f^{(1)} (x) = p {(x + 1)}^{p - 1}$ $f^{(2)} (x) = p (p - 1) {(x + 1)}^{p - 2} \Rightarrow f^{(m)} (x) = p (p - 1) . . . (p - m + 1) {(x + 1)}^{p - m}$ $= (- n) (- n - 1) . . . . (- n - m + 1) {(x + 1)}^{- n - m}$ $= {(- 1)}^{m} n (n + 1) . . . . (n + m - 1) {(x + 1)}^{- n - m} \Rightarrow$ $f (x) = \sum_{m = 0}^{n} \frac{f^{(m)} (0)}{m!} x^{m} = \sum_{m = 0}^{n} \frac{{(- 1)}^{m} n (n + 1) . . . (n + m - 1)}{m!} x^{m} \Rightarrow$ $a_{m} = \frac{{(- 1)}^{m} n (n + 1) . . . . (n + m - 1)}{m!}$
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