How i can find the sum of the series? $\binom{n}{1} + \binom{n}{2} + \cdots+ \binom{n}{\frac{n - 1}{2}} $ 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.
 A: 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...
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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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}

