Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was asked to find a closed formula for the sum

$$\sum_{k=0}^{n}\frac{1}{(k+1)(k+2)}\binom{n}{k}$$

could anyone give me an advice on how to get started?

share|improve this question
    
Hint: expand the binomal coefficient $\binom nk$ using factorials. –  V. Rossetto Feb 7 at 9:08
    
We get $$\sum_{k=0}^{n}\frac{n!}{(k+2)!(n-k)!}$$ –  Oria Gruber Feb 7 at 9:11
2  
That's correct. Now this almost looks like another binomial coefficient, doesn't it ? Try to write it as a binomial coefficient by multiplying it by something that depends on $n$ only. –  V. Rossetto Feb 7 at 9:14
    
gotcha :D thanks –  Oria Gruber Feb 7 at 9:17
add comment

5 Answers 5

up vote 2 down vote accepted

$$=\sum_{k=0}^{n}\frac{k!}{k!(k+1)(k+2)}\binom{n}{k} =\sum_{k=0}^{n}\frac{k!}{(k+2)!}\cdot\frac{n!}{k!(n-k)!} \\=\sum_{k=0}^{n}\frac{n!}{(k+2)!(n-k)!} \\=\frac{1}{(n+1)(n+2)}\sum_{k=0}^{n}\frac{(n+2)!}{(k+2)!(n-k)!} \\=\frac{1}{(n+1)(n+2)}\sum_{k=0}^{n}\binom{n+2}{k+2}$$ then you can complete using $(1+x)^n=\sum {{n}\choose{k}} x^k$

share|improve this answer
    
I think you meant $\sum \binom{n}{k}x^k$ and wrote $r$ by mistake, right? –  Oria Gruber Feb 7 at 9:22
    
yes @OriaGruber –  Semsem Feb 7 at 9:23
add comment

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $$ \mbox{Lets consider}\quad\fermi\pars{x}\equiv \sum_{k = 0}^{n}{x^{k + 2} \over \pars{k + 1}\pars{k + 2}}\,{n \choose k}\,,\quad \fermi\pars{0} =0\,,\quad\fermi\pars{1} = {\large ?} $$

$$ \fermi'\pars{x}= \sum_{k = 0}^{n}{x^{k + 1} \over k + 1}\,{n \choose k}\,,\qquad \fermi'\pars{0} = 0 $$

$$ \fermi''\pars{x}= \sum_{k = 0}^{n}x^{k}{n \choose k}=\pars{1 + x}^{n}\qquad\imp\qquad \fermi'\pars{x}={\pars{1 + x}^{n + 1} - 1\over n + 1} $$

$$ \imp\qquad\fermi\pars{x}= {1 \over n + 1}\,{\pars{1 + x}^{n + 2} - 1 \over n + 2} - {x \over n + 1} $$

$$ \fermi\pars{1} =\color{#66f}{\large\sum_{k = 0}^{n}{1 \over \pars{k + 1}\pars{k + 2}}\, {n \choose k} ={2^{n + 2} - n - 3 \over \pars{n + 1}\pars{n + 2}}} $$

share|improve this answer
    
Beautiful proof. –  MathFacts Jul 20 at 4:09
1  
@MathFacts I was wondering why no answer puts clearly and visible the final result. Thanks. –  Felix Marin Jul 20 at 7:41
    
I downloaded metmat-2003 from your site; Glad to know that Venezuela has a great mathematician. I am Brazilian and I have a lot to learn. –  MathFacts Jul 20 at 7:47
add comment

Hint
$$\sum_{k=0}^{n}\binom{n}{k}x^k = (1+x)^n$$ Integrate twice both rhs and lhs with respect to $x$ and when finished, plug $x=1$ in your result.

share|improve this answer
    
There is no need of integration here. –  V. Rossetto Feb 7 at 9:15
    
In "mathematic" house are many mansions (adaptation of John 14:2) –  Claude Leibovici Feb 7 at 9:19
    
I do agree with that, but one may enter into the house without keys to all mansions. –  V. Rossetto Feb 7 at 14:35
add comment

Sure - you know that $(1 + x)^n = \sum_{k=0} ^n \binom{n}{k} x^k$ from which it follows that $$ \frac{(1+x)^{n+1}}{n+1} = \int (1 + x)^n \, \mathrm{d}x = \int \sum_{k=0} ^n \binom{n}{k} x^k \, \mathrm{d}x = \sum_{k=0} ^n \binom{n}{k} \int x^k \, \mathrm{d}x = \sum_{k=0} ^n \binom{n}{k} \frac{x^{k+1}}{k+1} . $$ Based on what I've shown, you can iterate on this method, and allow $x$ to become a certain number, which should give you the closed form you are seeking.

share|improve this answer
add comment

Start with $(1+x)^n=\sum {{n}\choose{k}} x^k$ . If you integrate this twice wrt x you will get something close to what you are after.

share|improve this answer
    
You won by half a second ! Cheers. –  Claude Leibovici Feb 7 at 9:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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