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 $S_n$ is equal to:
- A$2^{\text{n}}-\text{n}-1$
- B$1-\frac{1}{2^{\text{n}}}$
- ✓$\text{n}-1-\frac{1}{2^{\text{n}}}$
- D${2^{\text{n}}}-1$
Answer: C.
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