If $\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}\frac{1+2+2^2+\ ...\text{ Sum to r terms}}{2^{\text{r}}},$ then Sn is equal to:
View full solution →- $2^{\text{n}}-\text{n}-1$
- $1-\frac{1}{2^{\text{n}}}$
- $\text{n}-1-\frac{1}{2^{\text{n}}}$
- ${2^{\text{n}}}-1$