Prove That the sum of the series: $\sum_{r=0}^{10} (-1)^{r}$ $\binom{10}{r}$ [$\frac{1}{2^{r}}+ \frac{3^{r}}{2^{2r}}+ \frac{7^{r}}{2^{3r}} +\frac{15^{r}}{2^{4r}}$ ....upto m terms =$\frac{2^{10m}-1}{2^{10m}(2^{10}-1)}$
I tried to expand this term but getting more and more complicated
Hint:
$$\sum_{r=0}^u\binom ur(-1)^r\left(\dfrac{2^n-1}{2^n}\right)^r=\sum_{r=0}^u\binom ur\left(\dfrac{1-2^n}{2^n}\right)^r=\left(1+\dfrac{1-2^n}{2^n}\right)^u=?$$
Here $n=1,2,\cdots,m$ and $u=10$