1
$\begingroup$

The Problem says that :-

If $\large \mathrm{N}$ is any positive integer, and $\delta_{n}=\large \int_{_{0}}^{^{N}} {{u} \choose {n}}{{N-u} \choose {N-n}} \textit{ du }$, then the value of

$\large \sum_{n=0}^{N}\delta_{n} = \\ \ \\ \ \\ (A). 0 \\ (B). N \\(C). N^2 \\ (D). \frac{1}{N} $ .

I was thinking about using hypergeometric distribution pmf in this problem but I don't see how I can apply that here. After using Fubini's theorem at the third step and after a few rearrangements, I am stuck at a queer expression which I don't know how to solve. Here is my approach :-

$\large \sum_{n=0}^{N}\delta_{n} \\ = \large \sum_{n=0}^{N} \large \int_{_{0}}^{^{N}} {{u} \choose {n}}{{N-u} \choose {N-n}} \textit{ du } \\ = \large \int_{_{0}}^{^{N}} \large \sum_{n=0}^{N} {{u} \choose {n}}{{N-u} \choose {N-n}} \textit{ du } \\ = \large \int_{_{0}}^{^{N}} \Bigg[\sum_{n=0}^{N} {{u} \choose {n}}{{N-u} \choose {N-n}} \Bigg] \textit{ du } \\ = \large \int_{_{0}}^{^{N}} \Bigg[\sum_{n=0}^{N} \frac{{{u} \choose {n}}{{N-u} \choose {k-n}}}{{{N} \choose {k}}} \frac{{{{N} \choose {k}}}(k-n)! (N-u-k+n)!}{(N-n)!(n-u)!} \Bigg] \textit{ du } \\ = \large \int_{_{0}}^{^{N}} {{N} \choose {k}} (N-u)!\Bigg[\sum_{n=0}^{N} \frac{{{u} \choose {n}}{{N-u} \choose {k-n}}}{{{N} \choose {k}}} \frac{(k-n)! (N-u-k+n)!}{(N-n)!(n-u)! (N-u)!} \Bigg] \textit{ du } \\ = \large \int_{_{0}}^{^{N}} {{N} \choose {k}} (N-u)!\Bigg[\sum_{n=0}^{N} \frac{{{u} \choose {n}}{{N-u} \choose {k-n}}}{{{N} \choose {k}}} \frac{1}{{{N-u} \choose {k-n}}(N-n)!(n-u)! (N-u)!} \Bigg] \textit{ du } \\ = \large \int_{_{0}}^{^{N}} \frac{{{N} \choose {k}}}{(N-u)!} \Bigg[\sum_{n=0}^{N} \frac{{{u} \choose {n}}{{N-u} \choose {k-n}}}{{{N} \choose {k}}} \frac{(N-u)!}{{{N-u} \choose {k-n}}(N-n)!(n-u)!} \Bigg] \textit{ du } \\ = \large \int_{_{0}}^{^{N}} \frac{{{N} \choose {k}}}{(N-u)!} \Bigg[\sum_{n=0}^{N} \frac{{{u} \choose {n}}{{N-u} \choose {k-n}}}{{{N} \choose {k}}} \frac{{N-u} \choose {n-u}}{{{N-u} \choose {k-n}}} \Bigg] \textit{ du } \\ =\large \int_{_{0}}^{^{N}} \frac{{{N} \choose {k}}}{(N-u)!} \Bigg[\sum_{n=0}^{N} \frac{{{u} \choose {n}} {{N-u} \choose {n-u}}}{{N} \choose {k}} \Bigg] \textit{ du } \longrightarrow (\star) $

and this is the $\star$, where I am stuck and cannot proceed further to solve this problem. I am not sure if my approach or thinking is wrong here. Any help is valuable and much appreciated. Also please kindly let me know if we can solve this problem with any other alternative approaches. Thanks in advance.


EDIT

Using the hint we have:-

$\large \sum_{n=0}^{N}\delta_{n} \\ = \large \sum_{n=0}^{N} \large \int_{_{0}}^{^{N}} {{u} \choose {n}}{{N-u} \choose {N-n}} \textit{ du } \\ = \large \int_{_{0}}^{^{N}} \large \sum_{n=0}^{N} {{u} \choose {n}}{{N-u} \choose {N-n}} \textit{ du } \\ = \large \int_{_{0}}^{^{N}} \Bigg[\sum_{n=0}^{N} {{u} \choose {n}}{{N-u} \choose {N-n}} \Bigg] \textit{ du } \\ = \large \int_{_{0}}^{^{N}} \Bigg[\sum_{n=0}^{N} \frac{{{N} \choose {n}}{{N} \choose {u}}^{-1}}{(u-n)! (n-u)!} \Bigg] \textit{ du } \\ = \large \int_{_{0}}^{^{N}} {{N} \choose {u}}^{-1} \Bigg[\sum_{n=0}^{N} \frac{{{N} \choose {n}}}{(u-n)! (n-u)!} \Bigg] \textit{ du } \\ = \large \int_{_{0}}^{^{N}} {{N} \choose {u}}^{-1} \Bigg[(\frac{1}{(u)!(-u)!} + \frac{1}{(u-N)!(N-u)!}) + (\frac{N}{(u-1)!(1-u)!}+ \frac{N}{(u-N+1)! (N-1-u)!}) + \cdots + \frac{{{N} \choose {\frac{(N+1)}{2}}}}{(u-\frac{(N+1)}{2})! (\frac{(N+1)}{2}-u)!} \Bigg] \textit{ du } \\ = \large \int_{_{0}}^{^{N}} {{N} \choose {u}}^{-1} \Bigg[(\frac{1}{\Gamma(u+1)\Gamma(1-u)} + \frac{1}{\Gamma(u-N+1)\Gamma(N-u+1)}) + (\frac{N}{\Gamma(u)\Gamma(2-u)}+ \frac{N}{\Gamma(u-N+2) \Gamma(N-u)}) + \cdots + \frac{{{N} \choose {\frac{(N+1)}{2}}}}{\Gamma(u-\frac{(N+1)}{2}+1) \Gamma(\frac{(N+1)}{2}-u +1)!} \Bigg] \textit{ du } \\ = \large \int_{_{0}}^{^{N}} {{N} \choose {u}}^{-1} \Bigg[(0+0) + (0+0) + \cdots + {{N} \choose {\frac{(N+1)}{2}}} \frac{(-1)^{u}}{\pi (u-\frac{(N+1)}{2})} \Bigg] \textit{ du } \\ = \large \int_{_{0}}^{^{N}} {{N} \choose {u}}^{-1} \Bigg[ {{N} \choose {\frac{(N+1)}{2}}} \frac{(-1)^{u}}{\pi (u-\frac{(N+1)}{2})} \Bigg] \textit{ du } \longrightarrow \star_{1}$

In $\star_{1}$, I am getting a weird expression which should result in $1$ but it doesn't. May I ask for any help regarding calculating the middle term of the sum?

Much thanks and appreciation for providing the hint.


Edit 2

I guess this should be the solution of the problem:-

We observe that the expression ${{u}\choose{n}}{{N-u}\choose{N-n}}$ becomes $1$ when $n=u$, hence $n \leq u < N$.

$\large \int_{_{0}}^{^{N}}\Bigg[\sum_{n=0}^{N}{{u}\choose{n}}{{N-u}\choose{N-n}} \Bigg] \textit{ du } \\ = \large \int_{_{0}}^{^{N}}\Bigg[ \underbrace{{u}\choose{0}}_{1}\underbrace{{N-u}\choose{N}}_{0}+\underbrace{{u}\choose{1}}_{u}\underbrace{{N-u}\choose{N-1}}_{0}+\cdots +\underbrace{{u}\choose{u-1}}_{u}\underbrace{{N-u}\choose{N-u+1}}_{0} +\underbrace{{u}\choose{u}}_{1}\underbrace{{N-u}\choose{N-u}}_{1}+\underbrace{{u}\choose{u+1}}_{0}\underbrace{{N-u}\choose{N-u-1}}_{N-u}+\cdots + \underbrace{{u}\choose{N}}_{0}\underbrace{{N-u}\choose{N-n}}_{\text{ positive term }} \Bigg] \textit{ du } \\ = \large \int_{_{0}}^{^{N}}1 \textit{ du } = N$

$\endgroup$
2
  • 2
    $\begingroup$ Hint: write ${{u} \choose {n}}{{N-u} \choose {N-n}} = {{N} \choose {n}}{{N} \choose {u}}^{-1}/\left((u-n)!(n-u)!\right)$ then use the reflection formula for the gamma function $\endgroup$ Feb 21 at 11:36
  • 1
    $\begingroup$ You can use the identity $\Gamma(z+1) = z\Gamma(z)$ for the latter. $\endgroup$
    – AxelT
    Feb 21 at 13:25

1 Answer 1

1
$\begingroup$

On the surface, since $\sum_{n=0}^N{{u} \choose {n}}{{N-u} \choose {N-n}}=1$ (there is only one nonezero element with a value of $1$), the answer is $B$, i.e. $\sum_{n=0}^N\delta_n=N$.

See this post too.

$\endgroup$
2
  • $\begingroup$ +1 For your answer. $\endgroup$
    – TopoSet32
    Feb 21 at 16:40
  • $\begingroup$ Happy to hear that this was helpful :-) $\endgroup$
    – Math-fun
    Feb 22 at 8:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .