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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
    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...

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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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    @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