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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^{nm} = \frac{1}{(1x)^{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 $\displaystyle \sum \limits_{n=m}^\infty \binom{n}{m} x^{nm} = 1+\binom{m+1}{m}x+\binom{m+2}{m}x^2+...=1+(m+1)x+ \frac{(m+2)(m+1)}{2} x^2+\frac{(m+3)(m+2)(m+1)}{3*2}x^3+...$ 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= nm. 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+m1}{i1}+\binom{i+m1}{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 $z=x$: $\displaystyle (1x)^\alpha=\sum_{k=0}^\infty\binom{\alpha}{k}(x)^k$ Let $\alpha=m1$: $\displaystyle \dfrac{1}{(1x)^{m+1}}=\sum_{k=0}^\infty\binom{m1}{k}(x)^k$ $\displaystyle \binom{m1}{k}=\dfrac{(m+1)\cdot((m+1)1)\dots((m+1)(k1))}{k!}=(1)^k\binom{m+k}{k}$ $\displaystyle \dfrac{1}{(1x)^{m+1}}=\sum_{k=0}^\infty\binom{m1}{k}(x)^k=\sum_{k=0}^\infty\binom{k+m}{k}x^k$ 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/(1x)^n = 1 + nx + [n(n+1)/2!]x^2 + ... 1/(1x)^(m+1) = 1 + (m+1)x + [(m+1)(m+2)/2!]x^2 + ...... 1/(1x)^(m+1) = 1 + (m+1)Cmx + (m+2)Cmx^2 + ......... nCm = n!/m!(nm)! (m+n)Cm=(m+n)!/m!n!= (m+1)(m+2)..(m+n)/n! 

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binomial, coefficients, identiti, identity 
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