Question

The sum of the series $\dfrac{1}{2} + \dfrac{3}{4} + \dfrac{7}{8} + \dfrac{{15}}{{16}} + ......$ up to low upon n term is:
A. $n - 1 + {2^{ - n}}$
B. ${2^{ - n}} - 1$
C. $n - 2$
D. None of these

Answer 1

Hint: In these types of questions you should try to modify the series so that they either become an A.P. or a G.P., and then apply the formula of the corresponding finite sum to find the sum of the given series.

Complete step-by-step solution
Let us consider the given series,
$\dfrac{1}{2} + \dfrac{3}{4} + \dfrac{7}{8} + \dfrac{{15}}{{16}} + ......$
We can write the given series as follows:
\[\left( {1 - \dfrac{1}{2}} \right) + \left( {1 - \dfrac{1}{4}} \right) + \left( {1 - \dfrac{1}{8}} \right) + \left( {1 - \dfrac{1}{{16}}} \right) + ......\]

Let us write this series in terms of the powers of 2 to get a clear picture of the series.
\[\left( {1 - \dfrac{1}{2}} \right) + \left( {1 - \dfrac{1}{{{2^2}}}} \right) + \left( {1 - \dfrac{1}{{{2^3}}}} \right) + \left( {1 - \dfrac{1}{{{2^4}}}} \right) + ......\]

Let us separate the 1’s from the above series and the fractions separately.
\[\left( {1 + 1 + 1 + .....n{\text{ terms}}} \right) - \left( {\dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} + .....n{\text{ terms}}} \right){\text{ }}......(1)\]

At this point we observe that the first bracket is the sum of n ones and the second bracket is a G.P, whose sum is given by; $S = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$ . For the series in the second bracket, the first term is $\dfrac{1}{2}$ and the common ratio is $\dfrac{1}{2}$.

Apply all this into (1) to get the sum of the given series;
\[
  S = n - \left( {\dfrac{{\dfrac{1}{2}\left( {1 - {{\left( {\dfrac{1}{2}} \right)}^n}} \right)}}{{1 - \dfrac{1}{2}}}} \right){\text{ }} \\
   \Rightarrow S = n - \left( {\dfrac{{\dfrac{1}{2}\left( {1 - \dfrac{1}{{{2^n}}}} \right)}}{{\dfrac{1}{2}}}} \right) \\
   \Rightarrow S = n - \left( {1 - \dfrac{1}{{{2^n}}}} \right){\text{ }} \\
   \Rightarrow S = n - 1 + \dfrac{1}{{{2^n}}} \\
   \Rightarrow S = n - 1 + {2^{ - n}} \\
 \]

Thus, we obtain the sum of the given series as $n - 1 + {2^{ - n}}$, which matches with option (A).

Note: It is important to make sure which type of series is given in the question, because the sum formula of both the infinite and finite A.P. or G.P. is different. Carefully find the first term and common ratio to substitute into the formulas known. Avoid making any calculation mistakes. Make yourself thorough with all the formulas of the sequence and series, and the meaning of the symbols used in them.