Answer
The Sum = $\frac{63}{128}$
Work Step by Step
Put i = 1 to 6 in $(\frac{1}{2})^{i + 1}$.
The Sum = $(\frac{1}{2})^{1 + 1}$ + $(\frac{1}{2})^{2 + 1}$ + $(\frac{1}{2})^{3 + 1}$ + $(\frac{1}{2})^{4 + 1}$ + $(\frac{1}{2})^{5 + 1}$ + $(\frac{1}{2})^{6 + 1}$
= $(\frac{1}{2})^{ 2}$ + $(\frac{1}{2})^{3}$ + $(\frac{1}{2})^{4}$ + $(\frac{1}{2})^{5}$ + $(\frac{1}{2})^{6}$ + $(\frac{1}{2})^{7}$
= $\frac{1}{4}$ + $\frac{1}{8}$ + $\frac{1}{16}$ + $\frac{1}{32}$ + $\frac{1}{64}$ + $\frac{1}{128}$
= $\frac{(32 + 16 + 8 + 4 + 2 +1)}{128}$
= $\frac{63}{128}$
