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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.

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  • $\begingroup$ About $a_n$, have a look to this amazing list of interesting answers to this question. By the way, this would give an answer to the question in your title which doesn't match your real question... $\endgroup$
    – Jean Marie
    Commented Mar 23, 2023 at 9:34

1 Answer 1

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

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

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