Looks like a reciprocal binomial

Algebra Level 4

Suppose we denote $\displaystyle a_n$ as the value of $\displaystyle \sum_{r=0}^n \dfrac1{^n C_r }$ , express $\displaystyle \sum_{r=0}^n \dfrac r{^n C_r}$ in terms of $n$ and/or $a_n$ .

Clarification :

$^n C_r$ denote the binomial coefficient, $\dbinom{n}{r} = \dfrac{n!}{r!(n-r)!}$ .

Practice for Bitsat here .

\left( n-1 \right) { a }_{ n }

\frac { 1 }{ 2 } n{ a }_{ n }

none of the above

n{ a }_{ n }

2 solutions

Ankit Nigam
Aug 31, 2015

Thanks for the solution sir. Its like JEE style, I did it in some generalized way...

$It\quad is\quad given\quad that\quad \\ { a }_{ n }=\frac { 1 }{ \left( \begin{matrix} n \\ 0 \end{matrix} \right) } +\frac { 1 }{ \left( \begin{matrix} n \\ 1 \end{matrix} \right) } +.........\frac { 1 }{ \left( \begin{matrix} n \\ n \end{matrix} \right) }$

Let S= $\sum _{ r=0 }^{ n }{ \frac { r }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } }$

$\therefore S=\frac { 0 }{ \left( \begin{matrix} n \\ 0 \end{matrix} \right) } +\frac { 1 }{ \left( \begin{matrix} n \\ 1 \end{matrix} \right) } +\frac { 2 }{ \left( \begin{matrix} n \\ 2 \end{matrix} \right) } ........+\frac { n }{ \left( \begin{matrix} n \\ n \end{matrix} \right) }$

or $S=\frac { n }{ \left( \begin{matrix} n \\ n \end{matrix} \right) } +\frac { n-1 }{ \left( \begin{matrix} n \\ n-1 \end{matrix} \right) } +\frac { n-2 }{ \left( \begin{matrix} n \\ n-2 \end{matrix} \right) } +......\frac { 0 }{ \left( \begin{matrix} n \\ 0 \end{matrix} \right) }$

$=\frac { n }{ \left( \begin{matrix} n \\ 0 \end{matrix} \right) } +\frac { n-1 }{ \left( \begin{matrix} n \\ 1 \end{matrix} \right) } +\frac { n-2 }{ \left( \begin{matrix} n \\ 2 \end{matrix} \right) } +......\frac { 0 }{ \left( \begin{matrix} n \\ n \end{matrix} \right) } \quad \left[ \because \left( \begin{matrix} n \\ r \end{matrix} \right) =\left( \begin{matrix} n \\ n-r \end{matrix} \right) \right]$

$\therefore 2S=\frac { n }{ \left( \begin{matrix} n \\ 0 \end{matrix} \right) } +\frac { n }{ \left( \begin{matrix} n \\ 1 \end{matrix} \right) } +\frac { n }{ \left( \begin{matrix} n \\ 2 \end{matrix} \right) } +.....\frac { n }{ \left( \begin{matrix} n \\ n \end{matrix} \right) }$

$\therefore 2S=n\left[ \frac { 1 }{ \left( \begin{matrix} n \\ 0 \end{matrix} \right) } +\frac { 1 }{ \left( \begin{matrix} n \\ 1 \end{matrix} \right) } +\frac { 1 }{ \left( \begin{matrix} n \\ 2 \end{matrix} \right) } +........\frac { 1 }{ \left( \begin{matrix} n \\ n \end{matrix} \right) } \right]$

$\therefore 2S=n\left[ { a }_{ n } \right]$

$\boxed { \therefore S=\frac { n{ a }_{ n } }{ 2 } }$

MY FIRST SOLUTION ON BRILLIANT :D

@Ankit Nigam , we really liked your comment, and have converted it into a solution. If you subscribe to this solution, you will receive notifications about future comments.

Brilliant Mathematics Staff - 5 years, 9 months ago

For quick solution, substitute n= 1 and match the given options..

Akhil Bansal - 5 years, 7 months ago

Chew-Seong Cheong
Apr 21, 2015

We can solve this by induction. Let $\space b_n = \displaystyle \sum_{r=0}^n{\frac{r}{^nC_r}}$ . And let us consider $\dfrac {b_n}{a_n}$ .

$\begin{array} {} \dfrac {b_0}{a_0} = \dfrac {\frac{0}{1}}{\frac{1}{1}} & = 0 & \Rightarrow b_0= (0)a_0 = \frac{1}{2}(0)a_0 \\ \dfrac {b_1}{a_1} = \dfrac {\frac{0}{1}+\frac{1}{1}}{\frac{1}{1}+\frac{1}{1}} & = \frac{1}{2} & \Rightarrow b_1= \frac{1}{2}a_1 = \frac{1}{2}(1)a_1 \\ \dfrac {b_2}{a_2} = \dfrac {\frac{0}{1}+\frac{1}{2}+\frac{2}{1}}{\frac{1}{1}+\frac{1}{2}+\frac{1}{1}} & = 1 & \Rightarrow b_2= a_2 = \frac{1}{2}(2)a_2 \\ \dfrac {b_3}{a_3} = \dfrac {\frac{0}{1}+\frac{1}{3}+\frac{2}{3}+\frac{3}{1}}{\frac{1}{1}+\frac{1}{3}+\frac{1}{3}+\frac{1}{1}} & = \frac{3}{2} & \Rightarrow b_3= \frac{3}{2} a_3 = \frac{1}{2}(3)a_3 \\ \dfrac {b_4}{a_4} = \dfrac {\frac{0}{1}+\frac{1}{4}+\frac{2}{6}+\frac{3}{4}+\frac{4}{1}}{\frac{1}{1}+\frac{1}{4}+\frac{1}{6}+\frac{1}{4}+\frac{1}{1}} & = 2 & \Rightarrow b_4 = 2 a_4 = \frac{1}{2}(4)a_4 \\ ... \end{array} $

$\Rightarrow b_n = \displaystyle \sum_{r=0}^n{\frac{r}{^nC_r}} = \boxed{\dfrac {1}{2}na_n}$

Moderator note:

This is not induction. You should show that the pattern always work for all integers as oppose to show that it appears to work for a small amount of integers.