2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I think stackexchange should have a way to send such suggestions privately. (Doesn't look good when it appears to the entire audience) $\endgroup$ – Kirthi Raman Mar 14 '12 at 11:18
  • $\begingroup$ I agree with that. $\endgroup$ – Tomarinator Mar 14 '12 at 11:21
  • $\begingroup$ None taken Kannappan Sampath. The policies can be refined as well when we are all participating. It doesn't hurt to suggest new ideas. $\endgroup$ – Kirthi Raman Mar 14 '12 at 12:06
10
$\begingroup$

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...

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ WolframAlpha disagrees. Also, the answer has a nice closed form of $2^{82}-1$. Did you forget to add the $\binom{83}{41}$ term? $\endgroup$ – JimmyK4542 Aug 10 '14 at 3:39
  • $\begingroup$ @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. $\endgroup$ – Felix Marin Aug 10 '14 at 4:09
  • $\begingroup$ @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. $\endgroup$ – Felix Marin Aug 10 '14 at 4:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.