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Question Number 197639 by mokys last updated on 25/Sep/23
show that (1/((1−z)^n )) = Σ_(k=1) ^∞ (((n+k−1)!)/(k!(n−1)!)) z^k
$${show}\:{that}\:\frac{\mathrm{1}}{\left(\mathrm{1}−{z}\right)^{{n}} }\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({n}+{k}−\mathrm{1}\right)!}{{k}!\left({n}−\mathrm{1}\right)!}\:{z}^{{k}} \\ $$
Commented by mr W last updated on 25/Sep/23
(1/((1−z)^n ))=Σ_(k=0) ^∞ (((k+n−1)),(( k)) ) z^k =Σ_(k=0) ^∞ (((k+n−1)),(( n−1)) ) z^k
$$\frac{\mathrm{1}}{\left(\mathrm{1}−{z}\right)^{{n}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{{k}+{n}−\mathrm{1}}\\{\:\:\:\:\:\:\:{k}}\end{pmatrix}\:{z}^{{k}} =\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{{k}+{n}−\mathrm{1}}\\{\:\:\:\:{n}−\mathrm{1}}\end{pmatrix}\:{z}^{{k}} \\ $$
Answered by mr W last updated on 25/Sep/23
f(x)=(1/((1−x)^n ))=(1−x)^(−n) f(0)=1 f^((1)) (x)=n(1−x)^(−(n+1)) ⇒f^((1)) (0)=n f^((2)) (x)=n(n+1)(1−x)^(−(n+2)) ⇒f^((2)) (0)=n(n+1) ... similarly f^((k)) (0)=n(n+1)...(n+k−1) acc. to taylor f(x)=Σ_(k=0) ^∞ ((f^((k)) (0))/(k!))x^k =Σ_(k=0) ^∞ ((n(n+1)...(n+k−1))/(k!))x^k =Σ_(k=0) ^∞ (((n+k−1)!)/(k!(n−1)!))x^k ✓
$${f}\left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{{n}} }=\left(\mathrm{1}−{x}\right)^{−{n}} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${f}^{\left(\mathrm{1}\right)} \left({x}\right)={n}\left(\mathrm{1}−{x}\right)^{−\left({n}+\mathrm{1}\right)} \:\Rightarrow{f}^{\left(\mathrm{1}\right)} \left(\mathrm{0}\right)={n} \\ $$$${f}^{\left(\mathrm{2}\right)} \left({x}\right)={n}\left({n}+\mathrm{1}\right)\left(\mathrm{1}−{x}\right)^{−\left({n}+\mathrm{2}\right)} \:\Rightarrow{f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)={n}\left({n}+\mathrm{1}\right) \\ $$$$… \\ $$$${similarly} \\ $$$${f}^{\left({k}\right)} \left(\mathrm{0}\right)={n}\left({n}+\mathrm{1}\right)…\left({n}+{k}−\mathrm{1}\right) \\ $$$${acc}.\:{to}\:{taylor} \\ $$$${f}\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{f}^{\left({k}\right)} \left(\mathrm{0}\right)}{{k}!}{x}^{{k}} \\ $$$$\:\:\:\:\:\:\:\:\:=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}\left({n}+\mathrm{1}\right)…\left({n}+{k}−\mathrm{1}\right)}{{k}!}{x}^{{k}} \\ $$$$\:\:\:\:\:\:\:\:\:=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({n}+{k}−\mathrm{1}\right)!}{{k}!\left({n}−\mathrm{1}\right)!}{x}^{{k}} \:\:\:\:\checkmark \\ $$

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