Find the value of $ \sum _{r=0} ^{2n} r ( ^{2n}C _r) ( \frac 1{r+2} ) $

Question

Find the value of $$ \sum _{r=0} ^{2n} r ( ^{2n}C _r ) ( \frac 1{r+2} )$$

---

In order to solve this I am trying to make the term(s) of the series independent of $r$. However I'm unable to solve this further: $$ \sum _{r=0} ^{2n} 2n (^{2n-1}C_{r-1})( \frac 1{r+2} ) $$
Any help would be appreciated. :)

Answer 1

$$\newcommand{\c}[2]{\binom{#1}{#2}} \newcommand{\r}[1]{\frac1{#1}} \newcommand{\s}[0]{\sum _{r=0} ^{2n}} \newcommand{\f}{\frac} \newcommand{\b}[1]{\left(#1\right)} \newcommand{\dx}{{\rm d}x}$$

$$\bbox[5pt,border:2px solid violet]{\sum _{r=0} ^{2n} 2n \c{2n-1}{r-1}\r{r+2}}$$

Method-I: Using: $$\bbox[border:2px solid blue,5pt]{\c nr=\frac nr\c{n-1}{r-1}}$$ $$\begin{align}\color{violet}{\sum _{r=0} ^{2n} 2n \c{2n-1}{r-1}\r{r+2}}&=\s \c{2n}r\f r{r+2}=\s \b{\c{2n}r-\f2{r+2}\c{2n}{r}}=4^n-2\color{indigo}{\s \r{r+2}\c{2n}{r}}\\\hline\color{indigo}{\s \r{r+2}\c{2n}{r}}&=\s \r{r+2}\cdot\f{(r+1)}{(2n+1)}\c{2n+1}{r+1}\\&=\r{2n+1}\s\c{2n+1}{r+1}-\s\r{(r+2)(2n+1)}\c{2n+1}{r+1}\\&=\r{2n+1}\s\c{2n+1}{r+1}-\r{(2n+1)(2n+2)}\s\c{2n+2}{r+2}\\&=\f{2^{2n+1}-1}{2n+1}-\f{4^{n+1}-(2n+3)}{(2n+1)(2n+2)}\\\hline\end{align}$$

$$\bbox[5pt,border:2px solid black]{\sum _{r=0} ^{2n} 2n \c{2n-1}{r-1}\r{r+2}= \f{2^{2 n+1} n^2-4^n n+4^n-1}{(n+1) (2 n+1)}}$$

---

Method-II: Using: $$\bbox[border:2px solid blue,5pt]{(1+x)^n=\sum_{r=0}^n\c nrr}$$

$$\sum _{r=0} ^{2n} 2n \c{2n-1}{r-1}\r{r+2}=2n\color{blue}{\sum _{r=0} ^{2n-1} \c{2n-1}{r}\r{r+3}}\tag{mind the -1!}$$

$$\color{blue}{\sum _{r=0} ^{2n-1} \c{2n-1}{r}\r{r+3}}=\sum _{r=0} ^{2n-1}\b{ \c{2n-1}{r}\f{x^{r+3}}{r+3}\Bigg|_{x=1}- \c{2n-1}{r}\f{x^{r+3}}{r+3}\Bigg|_{x=0}}\\=\int_0^1x^2(1+x)^{2n-1}\dx=\int_1^2x^{2n-1}(x-1)^2\dx=\int_1^2(x^{2n+1}-2x^{2n}+x^{2n-1})\dx\\=\b{\f{x^{2n+2}}{2n+2}-2\f{x^{2n+1}}{2n+1}+\f{x^{2n}}{2n}}_1^2=\b{\f{2^{2n+2}-1}{2n+2}-2\f{2^{2n+1}-1}{2n+1}+\f{2^{2n}-1}{2n}}$$

$$\bbox[5pt,border:2px solid black]{\sum _{r=0} ^{2n} 2n \c{2n-1}{r-1}\r{r+2}= \f{2^{2 n+1} n^2-4^n n+4^n-1}{(n+1) (2 n+1)}}$$

Answer 2

HINT:

$$r\binom{2n}r\cdot\frac1{r+2}=\frac{r(r+1)}{(2n+2)(2n+1)}\cdot\frac{(2n+2)!}{[(2n+2)-(r+2)]!\cdot(r+2)!}$$

$$=\frac{r(r+1)}{(2n+2)(2n+1)}\cdot\binom{2n+2}{r+2}$$

Now let $r(r+1)=(r+2)(r+1)+A(r+2)+B$

$r^2+r=r^2+r(3+A)+2+2A+B$

$\implies A+3=0\iff A=-3$

and $B+2A+2=0\iff B=-2A-2=4$

$$\implies(2n+2)(2n+1)r\cdot \binom{2n}r\cdot\frac1{r+2}=[(r+2)(r+1)+(-3)(r+2)+4]\binom{2n+2}{r+2}$$

$$=(r+2)(r+1)\binom{2n+2}{r+2}-3(r+2)\binom{2n+2}{r+2}+4\binom{2n+2}{r+2}$$

$$=(2n+2)(2n+1)\binom{2n}r-3(2n+1)\binom{2n+1}{r+1}+4\binom{2n+2}{r+2}$$

Finally, $\sum_{r=0}^{2n}\binom{2n}r=(1+1)^{2n}$

and $\sum_{r=0}^{2n}\binom{2n+1}{r+1}=(1+1)^{2n+1}-\binom{2n+1}0$

and $\sum_{r=0}^{2n}\binom{2n+2}{r+2}=(1+1)^{2n+2}-\binom{2n+2}0-\binom{2n+2}1$

Answer 3

Do you mean $\sum_{r=0}^{2n} \dfrac{r}{r+2} {2n \choose r}$? Maple says it's $${\frac {2\,{4}^{n}{n}^{2}-{4}^{n}n+{4}^{n}-1}{ \left( 2\,n+1 \right) \left( n+1 \right) }} $$

Answer 4

Here is an approach. Differentiating the identity

$$ S = \sum _{r=0} ^{2n} ( ^{2n}C _r) x^{r} = (1+x)^{2n}. $$

with respect to $x$ gives

$$ S'(x) = \sum _{r=0} ^{2n} r( ^{2n}C _r) x^{r-1}= 2n ( 1+x )^{2n-1} \longrightarrow (1).$$

Multiplying $(1)$ by $x^2$, integrating from $0$ to $1$ w.r.t. $x$ and then substituting $x=1$ gives the desired result

$$ \sum _{r=0} ^{2n} r( ^{2n}C _r) \frac{x^{r+2}}{r+2} = 2n\int_{0}^{1} x^2(1+x)^{2n-1} dx=\dots\,. $$

I let you do the integration and finish the problem.