It is currently Tue Jun 18, 2019 9:48 pm


All times are UTC [ DST ]




Post new topic Reply to topic  [ 2 posts ] 
Author Message
 Post subject: Binomial sum
PostPosted: Fri Jan 15, 2016 5:51 pm 
Administrator
Administrator
User avatar

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}$$

_________________
Imagination is much more important than knowledge.
Image


Top
Offline Profile  
Reply with quote  

 Post subject: Re: Binomial sum
PostPosted: 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. :)

_________________
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$


Top
Offline Profile  
Reply with quote  

Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 2 posts ] 

All times are UTC [ DST ]


Mathimatikoi Online

Users browsing this forum: No registered users and 1 guest


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
cron
Powered by phpBB® Forum Software © phpBB Group Color scheme created with Colorize It.
Theme created StylerBB.net