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Find the sum of the series when n is equal to 83?

$$\binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{\frac{n - 1}{2}} $$

I have got some idea that the trick to solve this particular problem is by using

$\dfrac{83-1}{2} =41$

But I am not getting how?

Thanks in advance.

  • 0
    I think stackexchange should have a way to send such suggestions privately. (Doesn't look good when it appears to the entire audience)2012-03-14
  • 0
    I agree with that.2012-03-14
  • 0
    None taken Kannappan Sampath. The policies can be refined as well when we are all participating. It doesn't hurt to suggest new ideas.2012-03-14

2 Answers 2

10

Hint:

  • $\displaystyle \sum_{r=0}^n \binom n r=2^n$

  • $\displaystyle \binom n r=\binom n {n-r}$

  • $n$ is odd.

Cook all of these...

1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Thanks to a comment of $\ds{\tt@JimmyK_{4542}}$, I found a missing term in a previous calculation. Indeed, the result turns out to be very simple:

\begin{align} &\color{#66f}{\large\sum_{k = 1}^{41}{83 \choose k}} =\half\bracks{% \sum_{k = 1}^{41}{83 \choose k} + \sum_{k = 1}^{41}{83 \choose 83 - k}} =\half\bracks{% \sum_{k = 1}^{41}{83 \choose k} + \sum_{k = -82}^{-42}{83 \choose -k}} \\[3mm]&=\half\bracks{% \sum_{k = 1}^{41}{83 \choose k} + \sum_{k = 82}^{42}{83 \choose k}} =\half\bracks{% \sum_{k = 0}^{83}{83 \choose k} - {83 \choose 0} - {83 \choose 83}} =\half\pars{2^{83} - 2} \\[3mm]&=\color{#66f}{\Large 2^{82} - 1} \end{align}

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    [WolframAlpha](http://www.wolframalpha.com/input/?i=sum+%2883+choose+k%29+from+k+%3D+1+to+41) disagrees. Also, the answer has a nice closed form of $2^{82}-1$. Did you forget to add the $\binom{83}{41}$ term?2014-08-10
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    @JimmyK4542 Indeed, I was expecting a nice result. Now, with your comment I'll check everything tomorrow because it's too late right now. Thanks.2014-08-10
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    @JimmyK4542 Indeed, the result was quite simple. I always have the idea of using the contour integral for the combinatoric number and it blinds me. You should publish your result and in that case I'll delete this one if they are similar. Thanks.2014-08-10