Binomial Summation

Let us assume that $\sum_{r=0}^{n}\frac{1}{\binom{n}{r}} = a$ Now we examine the sum $S_1 = \sum_{r=0}^{n}\frac{r}{\binom{n}{r}}$ and the sum $S_2 =\sum_{r=0}^{n}\frac{n-r}{\binom{n}{n-r}}$ clearly $S_1 = S_2$ and considering the identity $\binom{n}{r} = \binom{n}{n-r}$ it follows that $2S_1 = S_1+S_2 = \sum_{r=0}^{n}\frac{r}{\binom{n}{r}}+\sum_{r=0}^{n}\frac{n-r}{\binom{n}{n-r}} = \sum_{r=0}^{n}\frac{n}{\binom{n}{r}} = n\sum_{r=0}^{n}\frac{1}{\binom{n}{r}} = an$ and thus $S_1 = \frac{an}{2}$

We can write $a$ as $a = \dfrac{\color{#3D99F6}{1}}{\binom{n}{0}} + \dfrac{\color{#3D99F6}{1}}{\binom{n}{1}} + \dfrac{\color{#3D99F6}{1}}{\binom{n}{2}} + \cdots + \dfrac{\color{#3D99F6}{1}}{\binom{n}{n-2}} + \dfrac{\color{#3D99F6}{1}}{\binom{n}{n-1}} + \dfrac{\color{#3D99F6}{1}}{\binom{n}{n}}$

Let the required summation be $S$ .

$S = \dfrac{\color{#D61F06}{0}}{\binom{n}{0}} + \dfrac{\color{#D61F06}{1}}{\binom{n}{1}} + \dfrac{\color{#D61F06}{2}}{\binom{n}{2}} + \cdots + \dfrac{\color{#D61F06}{n-2}}{\binom{n}{n-2}} + \dfrac{\color{#D61F06}{n-1}}{\binom{n}{n-1}} + \dfrac{\color{#D61F06}{n}}{\binom{n}{n}}$

As we keep subtracting $a$ from $S$ , note that each coefficient reduces by 1.

$\begin{aligned} S - a & = \dfrac{\color{#D61F06}{-1}}{\binom{n}{0}} + \dfrac{\color{#D61F06}{0}}{\binom{n}{1}} + \dfrac{\color{#D61F06}{1}}{\binom{n}{2}} + \cdots + \dfrac{\color{#D61F06}{n-3}}{\binom{n}{n-2}} + \dfrac{\color{#D61F06}{n-2}}{\binom{n}{n-1}} + \dfrac{\color{#D61F06}{n-1}}{\binom{n}{n}} \\ S - 2a &= \dfrac{\color{#D61F06}{-2}}{\binom{n}{0}} + \dfrac{\color{#D61F06}{-1}}{\binom{n}{1}} + \dfrac{\color{#D61F06}{0}}{\binom{n}{2}} + \cdots + \dfrac{\color{#D61F06}{n-4}}{\binom{n}{n-2}} + \dfrac{\color{#D61F06}{n-3}}{\binom{n}{n-1}} + \dfrac{\color{#D61F06}{n-2}}{\binom{n}{n}} \\ \vdots & \qquad \quad \vdots \qquad \quad \vdots\qquad \quad \vdots\quad \qquad \quad \qquad \vdots \qquad \quad \vdots \quad \qquad \vdots\\ S - na &= \dfrac{\color{#D61F06}{-n}}{\binom{n}{0}} + \dfrac{\color{#D61F06}{-(n-1)}}{\binom{n}{1}} + \dfrac{\color{#D61F06}{-(n-2)}}{\binom{n}{2}} + \cdots + \dfrac{\color{#D61F06}{-2}}{\binom{n}{n-2}} + \dfrac{\color{#D61F06}{-1}}{\binom{n}{n-1}} + \dfrac{\color{#D61F06}{-0}}{\binom{n}{n}} \\ \end{aligned}$

Observe that the coefficients in $S - na$ are similar to the coefficients in $S$ . We can use the property of binomial coefficients , $\dbinom{n}{r} = \dbinom{n}{n-r}$ to see that $S - na$ is in fact equal to $-S$ .

$\begin{aligned} S - na &= \dfrac{\color{#D61F06}{-n}}{\binom{n}{n}} + \dfrac{\color{#D61F06}{-(n-1)}}{\binom{n}{n-1}} + \dfrac{\color{#D61F06}{-(n-2)}}{\binom{n}{n-2}} + \cdots + \dfrac{\color{#D61F06}{-2}}{\binom{n}{2}} + \dfrac{\color{#D61F06}{-1}}{\binom{n}{1}} + \dfrac{\color{#D61F06}{-0}}{\binom{n}{0}} \\ S - na & = - \left( \dfrac{\color{#D61F06}{n}}{\binom{n}{n}} + \dfrac{\color{#D61F06}{(n-1)}}{\binom{n}{n-1}} + \dfrac{\color{#D61F06}{(n-2)}}{\binom{n}{n-2}} + \cdots + \dfrac{\color{#D61F06}{2}}{\binom{n}{2}} + \dfrac{\color{#D61F06}{1}}{\binom{n}{1}} + \dfrac{\color{#D61F06}{0}}{\binom{n}{0}}\right) \\ S - na & = - \left( \dfrac{\color{#D61F06}{0}}{\binom{n}{0}} + \dfrac{\color{#D61F06}{1}}{\binom{n}{1}} + \dfrac{\color{#D61F06}{2}}{\binom{n}{2}} + \cdots + \dfrac{\color{#D61F06}{n-2}}{\binom{n}{n-2}} + \dfrac{\color{#D61F06}{n-1}}{\binom{n}{n-1}} + \dfrac{\color{#D61F06}{n}}{\binom{n}{n}} \right) \\ S - na & = -S\\ \end{aligned}$

Since $S - na = -S$ , we see that $\boxed{S = \dfrac{na}{2}}$ $_\square$

If $n$ is odd, we can rewrite $\sum _{ r=0 }^{ n }{ \left[ \frac { r }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } \right] }$ as $\sum _{ r=0 }^{ \frac { n-1 }{ 2 } }{ \left[ \frac { r }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } \right] +\left[ \frac { n-r }{ \left( \begin{matrix} n \\ n-r \end{matrix} \right) } \right] }$ , so we're pairing the first with the last, the second with the penultimate, until we get to the middle. As $n$ is odd, we're pairing an even number of terms.

As we know that $\left( \begin{matrix} n \\ r \end{matrix} \right) =\left( \begin{matrix} n \\ n-r \end{matrix} \right)$ , the sum can be simply written as $\sum _{ r=0 }^{ \frac { n-1 }{ 2 } }{ \left[ \frac { n }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } \right] }$ , which is equal to $n\cdot \sum _{ r=0 }^{ \frac { n-1 }{ 2 } }{ \left[ \frac { 1 }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } \right] }$

Basing on the same propertie as above, we note that $\frac { a }{ 2 } =\sum _{ r=0 }^{ \frac { n-1 }{ 2 } }{ \left[ \frac { 1 }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } \right] }$

Thus, $\sum _{ r=0 }^{ n }{ \left[ \frac { n }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } \right] } =\frac { na }{ 2 }$

If $n$ is even, we do the pairing terms process as follows:

$\sum _{ r=0 }^{ n }{ \left[ \frac { n }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } \right] } =\frac { n/2 }{ \left( \begin{matrix} n \\ { n }/{ 2 } \end{matrix} \right) } +\sum _{ r=0 }^{ \frac { n }{ 2 } -1 }{ \left[ \frac { r }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } \right] +\left[ \frac { n-r }{ \left( \begin{matrix} n \\ n-r \end{matrix} \right) } \right] } =\frac { n }{ 2\left( \begin{matrix} n \\ { n }/{ 2 } \end{matrix} \right) } +\sum _{ r=0 }^{ \frac { n }{ 2 } -1 }{ \left[ \frac { n }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } \right] } =n\left( \frac { 1 }{ 2\left( \begin{matrix} n \\ { n }/{ 2 } \end{matrix} \right) } +\sum _{ r=0 }^{ \frac { n }{ 2 } -1 }{ \left[ \frac { 1 }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } \right] } \right)$

And we can rewrite the last sum as

$n\left( \frac { 1 }{ 2 } \sum _{ r=0 }^{ n }{ \left[ \frac { 1 }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } \right] } \right) =\frac { na }{ 2 }$