Explanation

(d)
Let $S = 1 + 3 + 7 + 15 + ... + T_{n}$
$\Rightarrow \frac{S = 1 + 3 + 7 + ... + T_{n - 1} + T_{n}}{0 = 1 + 2 + 4 + 8 + ... - T_{n}}$
$\Rightarrow T_{n} = 1 + 2 + 4 + ...n$ terms
$= \frac{(2^{n} - 1)}{2 - 1} = 2^{n} - 1$
$\therefore\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + ... = \sum_{}^{}\frac{T_{n}}{2^{n}} = \sum_{}^{}\frac{2^{n} - 1}{2^{n}}$
$= \sum_{}^{}{(1 - 2^{- n})}$
$= n - \frac{\frac{1}{2}\left( 1 - \frac{1}{2^{n}} \right)}{\frac{1}{2}} = 2^{- n} + n - 1$