Question

The sum to $n$ terms of the series $\dfrac{1}{2} + \dfrac{3}{4} + \dfrac{7}{8} + \dfrac{15}{16} + \cdots$ is}

• A. $2^{n} - 1$
• B. $1 - 2^{n}$
• C. $n + 2^{n} - 1$
• D. $n - 1 + 2^{-n}$

Hint:

For series whose general term can be split, sum each part separately. The geometric part $\sum_{r=1}^n \left(\frac{1}{2}\right)^r = 1 - 2^{-n}$.

Answer 1

The Correct Option is D

Step 1: Conceptual Understanding:
Write the general term explicitly, then split the sum into simpler parts.
Step 2: Explanation in Detail:
The $r$-th term: $T_r = \dfrac{2^r - 1}{2^r} = 1 - \dfrac{1}{2^r}$.
\[S_n = \sum_{r=1}^n 1 - \sum_{r=1}^n \frac{1}{2^r} = n - \frac{\frac{1}{2}(1 - 2^{-n})}{1 - \frac{1}{2}} = n - (1 - 2^{-n}) = n - 1 + 2^{-n}.\] Step 3: Therefore, Stating the Final Answer
The sum to $n$ terms is $n - 1 + 2^{-n}$.