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Could you help me with a question I've tried to solve but stuck in the middle?

i need to calculate the coefficient $x^{2m}$ in every part of the algebraic identity $\frac{(1-x^2)^n}{(1-x)^n}=(1+x)^n$ in order to obtain the some sort of binomial identity: $\sum _{k=0} ^? ?? = \binom{n}{2m}$

what i did, i first verified that it holds true, and then got that the binom identity is $(1+x)^n=\sum_{i=o}^{n} \binom{n}{i}x^i$, but i don't think it's true. i also got that $x^{2m}=\binom {n}{m}$.

if relevant, i used the identity: $\frac{1}{(1-x)^n}=(1+x+x^2+x^3+...)^n=\sum_{k=0}^{\infty}D(n,k)x^k$ where $D(n,k) = \binom{n+k-1}{k}$

can you help me correct it please? how would you show it's true for m=2,n=5 for instance?

thank you very much for helping. i'm stuck on this one.

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  • $\begingroup$ $x^{2m}\color{red}{\neq} \binom {n}{2m}$ \begin{eqnarray*} \binom{n}{2m} = [x^{2m}]: (1+x)^{n} \end{eqnarray*} $\endgroup$ – Donald Splutterwit Sep 16 '17 at 13:06
  • $\begingroup$ sorry, $x^{2m}=\binom {n}{m}$ $\endgroup$ – BeginningMath Sep 16 '17 at 13:07
  • $\begingroup$ i can't develop the sum in order to obtain $\binom{n}{2m}$ at all $\endgroup$ – BeginningMath Sep 16 '17 at 13:11
  • $\begingroup$ @DonaldSplutterwit i've fixed my question. can you help me please solve it? i don't know how to obtain $\binom{n}{2m} $\endgroup$ – BeginningMath Sep 16 '17 at 13:20
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Another summation resulting in $\displaystyle\binom n{2m}$:

$$\sum_{j=2m}^n\binom {j-1}{2m-1}=\binom n{2m}$$

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\begin{eqnarray*} \binom{n}{2m} = [x^{2m}]: (1+x)^{n} \end{eqnarray*} \begin{eqnarray*} \frac{(1-x^2)^n}{(1-x)^n} = \left( \sum_{i=0}^{n} \binom{n}{i}(-1)^{i} x^{2i} \right) \left( \sum_{j=0}^{\infty} \binom{n+j-1}{j} x^{j} \right) \end{eqnarray*} We require the $x^{2m}$ terms from the above, so $2i+j=2m$ and $i$ can range over the values $0 \cdots m$. We have \begin{eqnarray*} [x^{2m}] : \frac{(1-x^2)^n}{(1-x)^n} = \sum_{i=0}^{n} \binom{n}{i}(-1)^{i} \binom{n+2(m-i)-1}{2(m-i)} \end{eqnarray*} and the combinatorial identity that you seek is \begin{eqnarray*} \sum_{i=0}^{n} \binom{n}{i}(-1)^{i} \binom{n+2(m-i)-1}{2(m-i)} = \binom{n}{2m}. \end{eqnarray*}

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  • $\begingroup$ we usually don't use the [] signs in discrete. could you explain what it means? is it floor value? $\endgroup$ – BeginningMath Sep 16 '17 at 13:43
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    $\begingroup$ \begin{eqnarray*} [x^{2m}]: (1+x)^{n} \end{eqnarray*} Reads as " The coeffieicent of ecks to the power of two emm in the function (open bracket) one plus ecks (close bracket) to the power of enn. $\endgroup$ – Donald Splutterwit Sep 16 '17 at 13:46

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