Finding sum of binomial series

Question

Find the sum of this series.

I tried writing the general term of this term. I succeeded in it but the sum inside the bracket depends on r which makes more problem to get overall sum. I know that this uses the concept of binomial theorem and binomial coefficients but I am not able to break through it to get the desired concept to use.

Answer 1

First, by the binomial theorem, $\sum_{r=0}^n (-1)^r\binom{n}{r}x^r =(1-x)^n $.

Second, the series on the right has terms $\dfrac{(2^j-1)^r}{2^{jr}} =\left(\dfrac{2^j-1}{2^j}\right)^r =(1-2^{-j})^r $ for $j = 1$ to $m$ so it is $\sum_{j=1}^{m} (1-2^{-j})^r $.

Note: In my original answer, I had 0 to m-1. This is a correction.

Therefore the sum is

$\begin{array}\\ \sum_{r=0}^n (-1)^r\binom{n}{r}\sum_{j=1}^{m} (1-2^{-j})^r &=\sum_{j=1}^{m}\sum_{r=0}^n (-1)^r\binom{n}{r} (1-2^{-j})^r\\ &=\sum_{j=1}^{m}(1-(1-2^{-j}))^n\\ &=\sum_{j=1}^{m}(2^{-j})^n\\ &=\sum_{j=1}^{m}(2^{-n})^j\\ &=2^{-n}\dfrac{1-2^{-mn}}{1-2^{-n}} \qquad\text{since }\sum_{j=1}^{m}x^j =x\dfrac{1-x^m}{1-x} \text{ with } x=2^{-n}\\ \end{array} $

Answer 2

Hint:

$$=\sum_{m=1}^\infty\left(\sum_{r=0}^n\binom nr\left(\dfrac{1-2^m}{2^m}\right)^r\right)$$