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 Post subject: Binomial sumPosted: Fri Jan 15, 2016 5:51 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Let $n \in \mathbb{N}$. Evaluate (in a closed form) the following sums:

$$\begin{matrix} {\text (a)} \; \displaystyle \sum_{k=0}^{n}\binom{2n}{2k}&&&&&&& \text{(b)} \; \displaystyle \sum_{k=0}^{n}\binom{3n}{3k} \end{matrix}$$

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 Post subject: Re: Binomial sumPosted: Wed Apr 06, 2016 7:43 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
For the second sum let $\omega=e^{2\pi i /3}$ be a third root of unity. The binomial expansion tells us

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

Subbing $x$ with $1, \omega, \omega^2$ respectively at the binomial expansion we get:

$$\begin{matrix} \displaystyle \sum_{k=0}^{n}\binom{n}{k}=2^n & , &\displaystyle \left ( 1+\omega \right )^n =\sum_{k=0}^{n}\binom{n}{k}\omega^n &, &\displaystyle \left ( 1+\omega^2 \right )^n = \sum_{k=0}^{n} \binom{n}{k}\omega^{2k} \end{matrix}$$

Adding the above three equations together we get that:

$$\sum_{k=0}^{n}\binom{n}{k}\left [ 1+\omega^k +\omega^{2k} \right ]=2^n + \left ( 1+\omega \right )^n + \left ( 1+\omega^2 \right )^n$$

We note that $1+\omega^k +\omega^{2k}=3$ when $k$ is divisible by $3$ and $0$ othewise. Hence:

$$\sum_{m=0}^{\left \lfloor n/3 \right \rfloor}\binom{n}{3m}= \frac{1}{3}\left [ 2^n + \left ( 1+\omega \right )^n+ \left ( 1+\omega^2 \right )^n \right ]$$

Subbing $n$ with $3n$ and doing the calculations we finally get that:

$$\sum_{k=0}^{n}\binom{3n}{3k}= \frac{1}{3}\left [ 8^n + 2(-1)^n \right ]$$

The first sum is much easier and I leave it to someone else.

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