Solve ${a_n} = \sum\limits_{r = 0}^n {\frac{1}{{{}^n{C_r}}}} = \frac{1}{{{}^n{C_0}}} + \frac{1}{{{}^n{C_1}}} +\ldots+ \frac{1}{{{}^n{C_n}}}$

Question

If $${a_n} = \sum\limits_{r = 0}^n {\frac{1}{{{}^n{C_r}}}} = \frac{1}{{{}^n{C_0}}} + \frac{1}{{{}^n{C_1}}} + \frac{1}{{{}^n{C_2}}} + \ldots + \frac{1}{{{}^n{C_n}}},$$ then find the value of $$\sum\limits_{r = 0}^n {\frac{r}{{{}^n{C_r}}}}$$ in terms of $a_n$. (Where ${{}^n{C_r}}$ is $\frac{{n!}}{{r!\left( {n - r} \right)!}}$.)

Not able to solve it as ${{}^n{C_r}}$ comes in the denominator.

Answer 1

$S=\sum\limits_{r = 0}^n {\frac{r}{{{}^n{C_r}}}}=\frac{0}{{{}^n{C_0}}} + \frac{1}{{{}^n{C_1}}} + \frac{2}{{{}^n{C_2}}} + \ldots + \frac{n}{{{}^n{C_n}}}$

Write same series after Reversing

$S=\frac{n}{{{}^n{C_n}}} + \frac{n-1}{{{}^n{C_{n-1}}}} + \frac{n-2}{{{}^n{C_{n-2}}}} + \ldots + \frac{0}{{{}^n{C_0}}}$

Now Add both series and use ${n\choose r} ={n\choose n-r} $

$2S=\frac{0+n}{{{}^n{C_0}}} + \frac{1+n-1}{{{}^n{C_1}}} + \frac{2+n-2}{{{}^n{C_2}}} + \ldots + \frac{n+0}{{{}^n{C_n}}}$

$\implies$

$2S=\frac{n}{{{}^n{C_0}}} + \frac{n}{{{}^n{C_1}}} + \frac{n}{{{}^n{C_2}}} + \ldots + \frac{n}{{{}^n{C_n}}}$

Hence $S=\dfrac{n}{2}\bigg(\frac{1}{{{}^n{C_0}}} + \frac{1}{{{}^n{C_1}}} + \frac{1}{{{}^n{C_2}}} + \ldots + \frac{1}{{{}^n{C_n}}}\bigg)$