My Math Forum Identity with binomial coefficients

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 March 10th, 2016, 04:59 AM #1 Newbie   Joined: Dec 2013 Posts: 4 Thanks: 0 Identity with binomial coefficients I have to show that $\displaystyle \sum \limits_{n=m}^\infty \binom{n}{m} x^{n-m} = \frac{1}{(1-x)^{m+1}}$ for $\displaystyle m \geq 0$ and $\displaystyle \forall x \in \mathbb{C}$ with $\displaystyle \|x\| <1$. If I compute the left side I get $\displaystyle \sum \limits_{n=m}^\infty \binom{n}{m} x^{n-m} = 1+\binom{m+1}{m}x+\binom{m+2}{m}x^2+...=1+(m+1)x+ \frac{(m+2)(m+1)}{2} x^2+\frac{(m+3)(m+2)(m+1)}{3*2}x^3+...$ but it doesn't help? Do you know what formula is needed to show the identity?
 March 10th, 2016, 06:35 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 I would not expand the binomial like this. Instead change the index: let i= n-m. Then n= i+ m and when n= m, i= 0. The sum becomes $\displaystyle \sum_{i=0}^\infty \begin{pmatrix}i+ m \\ m \end{pmatrix} x^i$. Now, do you know a "formula" for simplifying $\displaystyle \begin{pmatrix}i+ m \\ m \end{pmatrix}$. Last edited by greg1313; March 10th, 2016 at 01:38 PM.
 March 10th, 2016, 07:24 AM #3 Newbie   Joined: Dec 2013 Posts: 4 Thanks: 0 Thank you. Then I have $\displaystyle \sum \limits_{i=0}^\infty \binom{i+m}{m}x^i= \sum \limits_{i=0}^\infty \binom{i+m}{i}x^i= ..?$ I know that $\displaystyle \sum \limits_{i=0}^\infty \binom{\alpha}{i}x^i=(1+x)^\alpha$ but my $\displaystyle \alpha$ in the sum above still depends on i which does not allow applying this equation. $\displaystyle ?...= \sum \limits_{i=0}^\infty ( \binom{i+m-1}{i-1}+\binom{i+m-1}{i} ) x^i$ doesn't help either or what formula did you mean? Last edited by hubolmen; March 10th, 2016 at 07:27 AM.
 March 10th, 2016, 01:02 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Start with the binomial series: $\displaystyle (1+z)^\alpha=\sum_{k=0}^\infty\binom{\alpha}{k}z^k$ Let $z=-x$: $\displaystyle (1-x)^\alpha=\sum_{k=0}^\infty\binom{\alpha}{k}(-x)^k$ Let $\alpha=-m-1$: $\displaystyle \dfrac{1}{(1-x)^{m+1}}=\sum_{k=0}^\infty\binom{-m-1}{k}(-x)^k$ $\displaystyle \binom{-m-1}{k}=\dfrac{-(m+1)\cdot(-(m+1)-1)\dots(-(m+1)-(k-1))}{k!}=(-1)^k\binom{m+k}{k}$ $\displaystyle \dfrac{1}{(1-x)^{m+1}}=\sum_{k=0}^\infty\binom{-m-1}{k}(-x)^k=\sum_{k=0}^\infty\binom{k+m}{k}x^k$ Last edited by greg1313; March 10th, 2016 at 01:37 PM.
 March 22nd, 2016, 01:05 PM #5 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 I see greg1313 has already done this. If you want to see it slightly differently: 1/(1-x)^n = 1 + nx + [n(n+1)/2!]x^2 + ... 1/(1-x)^(m+1) = 1 + (m+1)x + [(m+1)(m+2)/2!]x^2 + ...... 1/(1-x)^(m+1) = 1 + (m+1)Cmx + (m+2)Cmx^2 + ......... nCm = n!/m!(n-m)! (m+n)Cm=(m+n)!/m!n!= (m+1)(m+2)..(m+n)/n!

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