# Binomial Expansion

• Dec 14th 2013, 06:04 AM
srirahulan
Binomial Expansion
$\displaystyle C{_0}^{2}+C{_1}^{2}+C{_2}^{2}+...............+C{_0 }^{2}=\frac{2n!}{n!^2}$

$\displaystyle C{_0}^{2} - C{_1}^{2}+C{_2}^{2} - C{_3}^{2}+..........+(-1)C{_n}^{n}=\frac{(-1)^{\frac{n}{2}}n!}{n/2)!^2}$ ---->>> (n even number)
=0 --------------------------------------------------------->>>>(n odd number)

i use binomial expansion theorem but i cannot get the answer like this.....please give me hint for these problems>>>><<<<
• Dec 14th 2013, 12:04 PM
SlipEternal
Re: Binomial Expansion
Your notation is not at all standard. I will help you rewrite it. Press Reply With Quote to see how I did it.

$\displaystyle \binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \cdots + \binom{n}{n}^2 = \binom{2n}{n} = \dfrac{(2n)!}{(n!)^2}$

$\displaystyle \binom{n}{0}^2 - \binom{n}{1}^2 + \binom{n}{2}^2 - \binom{n}{3}^2 \pm \cdots + (-1)^n\binom{n}{n}^2 = \begin{cases}\dfrac{(-1)^{n/2}n!}{[(n/2)!]^2} & n\text{ is even} \\ 0 & n\text{ is odd}\end{cases}$

For the first one,

$\displaystyle \sum_{k=0}^{2n}\binom{2n}{k}x^k = (1+x)^{2n} = (1+x)^n(1+x)^n = \left(\sum_{k=0}^n \binom{n}{k}x^k \right)^2$

For the coefficient of $\displaystyle x^n$, you have $\displaystyle \binom{2n}{n}$ on the LHS. On the RHS, you need to multiply out first. Each term will be of the following form: $\displaystyle \binom{n}{k_1}x^{k_1}\binom{n}{k_2}x^{k_2} = \binom{n}{k_1}\binom{n}{k_2}x^{k_1+k_2}$. So, you want to add up all terms where $\displaystyle k_1+k_2 = n$. That means $\displaystyle k_2 = n-k_1$. So,

$\displaystyle \binom{2n}{n} = \binom{n}{0}\binom{n}{n-0} + \binom{n}{1}\binom{n}{n-1} + \cdots + \binom{n}{n-1} \binom{n}{n-(n-1)} + \binom{n}{n}\binom{n}{n-n}$

Since $\displaystyle \binom{n}{k} = \binom{n}{n-k}$ for all $\displaystyle k$, you can rewrite the RHS as:

$\displaystyle \binom{n}{0}\binom{n}{0} + \cdots + \binom{n}{n}\binom{n}{n} = \binom{n}{0}^2 + \cdots + \binom{n}{n}^2$
• Dec 14th 2013, 12:13 PM
SlipEternal
Re: Binomial Expansion
For the second one, do something similar with $\displaystyle \sum_{k=0}^n (-1)^k\binom{n}{k}x^{2k} = (1-x^2)^n = (1+x)^n(1-x)^n = \left(\sum_{k=0}^n \binom{n}{k}x^k\right)\left(\sum_{k=0}^n (-1)^k\binom{n}{k}x^k\right)$. Again, find the coefficient of $\displaystyle x^n$ on each side.