oh...my questions are challenge enough..wish someone can answer me this binomial theory...i really sucks at this..

Show that,

$\displaystyle

\sum_{k=0}^{n}\binom{k+n}{k}{2}^{-k}=2^n

$

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- Sep 17th 2009, 06:35 PMmanusformbinomials...
oh...my questions are challenge enough..wish someone can answer me this binomial theory...i really sucks at this..

Show that,

$\displaystyle

\sum_{k=0}^{n}\binom{k+n}{k}{2}^{-k}=2^n

$ - Sep 17th 2009, 09:02 PMBingk
Did you copy that correctly?

If it's supposed to be $\displaystyle \sum_{k=0}^{n+k}\binom{k+n}{k}{2}^{-k}=2^n$

This was already shown $\displaystyle \sum_{k=0}^{n}\binom{n}{k}=2^n$

So, $\displaystyle \sum_{k=0}^{n+k}\binom{n+k}{k}=2^{n+k}$

This gives us $\displaystyle \sum_{k=0}^{n+k}\binom{n+k}{k}2^{-k}=2^{n}$ - Sep 19th 2009, 03:57 AMOpalg
Sorry to have to say so, but the expression in red makes no sense (nor does the rest of that argument). if k is the summation index, then it cannot also occur as part of one of the limits of summation. The summation index is a variable, and the limits of summation must be constants.

The left side of that identity is the coefficient of $\displaystyle x^n$ in the expression $\displaystyle \sum_{k=0}^{n}\bigl(x+\tfrac12\bigr)^{n+k}$. But that is a geometric series, with sum $\displaystyle \frac{\bigl(x+\tfrac12\bigr)^{2n+1} - \bigl(x+\tfrac12\bigr)^n}{x-\tfrac12}$. To find the coefficient of $\displaystyle x^n$ there, write $\displaystyle \frac1{x-\tfrac12} = -2(1-2x)^{-1} = -2\bigl(1+2x+(2x)^2+(2x)^3+\ldots\bigr)$.

Picking out the coefficient of $\displaystyle x^n$ in $\displaystyle -2\bigl(\bigl(x+\tfrac12\bigr)^{2n+1} - \bigl(x+\tfrac12\bigr)^n\bigr) \bigl(1+2x+(2x)^2+(2x)^3+\ldots\bigr)$, you find that it is

$\displaystyle \begin{aligned} -2\biggl(&{2n+1\choose n}2^{-n-1} + {2n+1\choose n-1}2^{-n-1} + {2n+1\choose n}2^{-n-1} + \ldots + {2n+1\choose 0}2^{-n-1} \\ & - 1 - {n\choose 1}- \ldots - {n\choose n}\biggr).\end{aligned}$

But in the second line of that expression, the sum of all the binomial coefficients $\displaystyle \textstyle {n\choose k}$ is $\displaystyle 2^n$. In the first line, the sum of half of the binomial coefficients $\displaystyle \textstyle {2n+1\choose k}$ (with k going from 0 to n) is $\displaystyle \tfrac12\cdot2^{2n+1} = 2^{2n}$. That gives the value for the coefficient of $\displaystyle x^n$ as $\displaystyle -2(2^{n-1} - 2^n) = 2^n.$ - Sep 20th 2009, 07:17 PMBingk
Yup, you're absolutely right, I completely missed that :) ... sorry.