I'm trying to find the coefficient of $x^{2m}$ from the both sides of the following equality:

$$\frac{(1-x^2)^n}{(1-x)^n} = (1+x)^n$$

For the right side of equality I've found it as follow:

$$[x^{2m}] (1+x)^n=\binom{n}{2m}$$

With the left side of equation I have a problem: $$[x^{2m}] (1-x^2)^n(1-x)^{-n}=?$$

Also, according to previous computations of coefficient, is needed to build the following equality of the sums of binomial-coefficients:

$$\sum_{k=0}^? ??=\binom{n}{2m}$$

Can some one please explain how to get it?

For example if given that $n=5$ and $m=2$:

$$\sum_{k=0}^n \binom{k}{k} = \binom{n}{2m}$$

$$\sum_{k=0}^5 \binom{k}{k} = \binom{5}{4}$$

$$\binom{0}{0} + \binom{1}{1} + \binom{2}{2} + \binom{3}{3} + \binom{4}{4} = \binom{5}{4}$$

$$1+1+1+1+1 = 5$$

It seems to work, but what if given $n=5$ and $m=3$? Then I'm in trouble...

What will be the correct sigma notation for this case?

Regards.

• @graydad, I think the $[x^{2m}](...)$ means the coefficient of $x^{2m}$ in $(...)$. So in this case, the OP is simply using the binomial theorem. – TravisJ Jun 15 '15 at 18:52
• I've put $[x^{2m}]$ just to denote that I'm looking for it's coefficient in $(a+x)^n$ – Mike Jun 15 '15 at 18:53
• What is wrong with factoring the LHS so that it it $((1-x)(1+x))^{n}$ and cancelling. Then you know the LHS and the RHS are the same, and you know the coefficient of $x^{2m}$ in the RHS... so it must be the same on the LHS? – TravisJ Jun 15 '15 at 18:54
• @TravisJ Yes, it must be the same as on the RHS but I need to show it, so I need to find the coefficient on the LHS as well. To my shame I have problem with factoring. and what to do with negative power of n.. Can you explain a bit how to solve the LHS? – Mike Jun 15 '15 at 19:01
• Why not just say $\frac{(1-x^{2})^{n}}{(1-x)^{n}}=\frac{((1-x)(1+x))^{n}}{(1-x)^{n}}=(1+x)^{n} \frac{(1-x)^{n}}{(1-x)^{n}}=(1+x)^{n}$? – TravisJ Jun 15 '15 at 19:11

If you want to write $\binom{n}{m}$ as a sum of binomial coefficients, you can do it in the following way (assuming $0\leq m< n$). We will repeatedly apply Pascal's identity: $\binom{n}{m}=\binom{n-1}{m}+\binom{n-1}{m-1}$.

\begin{align*} \binom{n}{m} &= \binom{n-1}{m}+\binom{n-1}{m-1} \\ &= \binom{n-1}{m}+\binom{n-2}{m-1}+\binom{n-2}{m-2} \\ &= \binom{n-1}{m}+\binom{n-2}{m-1}+\binom{n-3}{m-2}+\binom{n-3}{m-3} \\ &\vdots \\ &=\sum_{i=0}^{m}\binom{n-(i+1)}{m-i} \end{align*}