If the summation of $\frac{1}{^nC_r}$ is $a_n$, find the summation of $\frac{r}{^nC_r}$ in terms of $a_n$

Question

My attempt to solve this question:

1. Write $a_n=\Sigma^n_{r=0}\frac{1}{^nC_r}$ as => $a_n=\frac{1}{^nC_0}+\frac{x}{^nC_1}+\frac{x^2}{^nC_2}...+\frac{x^n}{^nC_n}$
2. Differentiate $a_n$ with respect to x, obtaining: $a_n=\frac{1}{^nC_1}+\frac{2x}{^nC_2}+\frac{3x^2}{^nC_3}...+\frac{nx^{n-1}}{^nC_n}$
3. Replace $x=1$ in $\frac{d(a_n)}{dx}$, but now I don't have $a_n$ as a function of x.

So I guess my basic question is which binomial expansion has $^nC_r$ in the denominator, which I can use to write $a_n$ as a function of x?

Answer 1

The key here is to try to rewrite $\frac{r}{^nC_r}$ in terms of the sequence $\frac{1}{^nC_r}$. We find that

$$\frac{r}{^nC_r} = \frac{r\cdot r!(n-r)!}{n!} = \frac{(r+1 - 1)\cdot r!(n-r)!}{n!}\\ = \frac{(r+1)!(n-r)!}{n!} - \frac{r!(n-r)!}{n!} = (n+1)\cdot \frac{1}{^{n+1}C_{r+1}} - \frac{1}{^nC_r}$$

Now you can try to sum the equality $\frac{r}{^nC_r} = (n+1)\cdot \frac{1}{^{n+1}C_{r+1}} - \frac{1}{^nC_r}$ from $r=0$ to $r=n$ and use the definition of $a_n$ on the two terms you get on the right hand side to simplify.