sum of the Series $\sum^{n}_{r=0}(-1)^r\binom{n}{r}\left[\frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+........\bf{m\; terms}\right]$

Question

The sum of the Series $\displaystyle \sum^{n}_{r=0}(-1)^r\binom{n}{r}\left[\frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+........\bf{m\; terms}\right]$

$\bf{My\; Try::}$Let $$\displaystyle S= \sum^{n}_{r=0}(-1)^r\binom{n}{r}\left[\frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+........\bf{m\; terms}\right]$$

So $$S=\sum^{n}_{r=0}(-1)^r\binom{n}{r}\left[\left(1-\frac{1}{2}\right)^r+\left(1-\frac{1}{2^2}\right)^r+\left(1-\frac{1}{2^3}\right)^r+.....+\left(1-\frac{1}{2^m}\right)^r\right]$$

So $$S=\sum^{n}_{r=0}(-1)^r\binom{n}{r}\left(1-\frac{1}{2}\right)^r+\sum^{n}_{r=0}(-1)^r\binom{n}{r}\left(1-\frac{1}{2^2}\right)^r+.....+\sum^{n}_{r=0}(-1)^r\binom{n}{r}\left(1-\frac{1}{2^m}\right)^r$$

Now Using $$\bullet\; (1-x)^n=\sum^{n}_{r=0}(-1)^r\binom{n}{r}x^r.$$

So we get $$S=\frac{1}{2^n}+\frac{1}{2^{2n}}+........+\frac{1}{2^{mn}} = \frac{\frac{1}{2^n}-\frac{1}{2^{mn+n}}}{1-\frac{1}{2^n}}=\frac{2^{mn}-1}{2^{mn}\cdot(2^n-1)}$$

My Question is can we solve it any other way, If yes then plz explain here

Thanks