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March 10th, 2016, 04:59 AM   #1
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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?
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March 10th, 2016, 06:35 AM   #2
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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.
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March 10th, 2016, 07:24 AM   #3
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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.
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March 10th, 2016, 01:02 PM   #4
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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.
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March 22nd, 2016, 01:05 PM   #5
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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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