Question

The value of $\dfrac{3}{4} + \dfrac{{15}}{{16}} + \dfrac{{63}}{{64}} + ......$ up to n terms is
${\text{A}}{\text{. }}n - \dfrac{{{4^n}}}{3} - \dfrac{1}{3}$
${\text{B}}{\text{. }}n + \dfrac{{{4^{ - n}}}}{3} - \dfrac{1}{3}$
${\text{C}}{\text{. }}n + \dfrac{{{4^n}}}{3} - \dfrac{1}{3}$
${\text{D}}{\text{. }}n - \dfrac{{{4^{ - n}}}}{3} + \dfrac{1}{3}$

Answer 1

Hint: Write the series $\dfrac{3}{4} + \dfrac{{15}}{{16}} + \dfrac{{63}}{{64}} + ......$ as \[\left( {1 - \dfrac{1}{4}} \right) + \left( {1 - \dfrac{1}{{16}}} \right) + \left( {1 - \dfrac{1}{{64}}} \right) + .......\] to n terms, then solve the question.

Complete step-by-step answer:
We have been given in the question, the series $\dfrac{3}{4} + \dfrac{{15}}{{16}} + \dfrac{{63}}{{64}} + ......$ up to n terms.
To find – the sum of the given series.
We can write the given series in the form-
\[\left( {1 - \dfrac{1}{4}} \right) + \left( {1 - \dfrac{1}{{16}}} \right) + \left( {1 - \dfrac{1}{{64}}} \right) + .......\] to n terms.
Solving it further-
\[
  \left( {1 - \dfrac{1}{4}} \right) + \left( {1 - \dfrac{1}{{16}}} \right) + \left( {1 - \dfrac{1}{{64}}} \right) + ....... \\
   = (1 + 1 + 1 + ... + 1) - \dfrac{1}{4}\left( {1 + \dfrac{1}{4} + \dfrac{1}{{{4^2}}} + .......} \right) \\
   = n - \dfrac{1}{4}.\dfrac{{1 - {{\left( {\dfrac{1}{4}} \right)}^n}}}{{1 - \dfrac{1}{4}}} \\
   = n - \dfrac{1}{4}.\dfrac{{(1 - {4^{ - n}})}}{{\dfrac{3}{4}}} \\
   = n - \dfrac{1}{3}(1 - {4^{ - n}}) \\
   = n + \dfrac{{{4^{ - n}}}}{3} - \dfrac{1}{3} \\
\] {Since, the sum \[\left( {1 + \dfrac{1}{4} + \dfrac{1}{{{4^2}}} + .......} \right)\]=\[\dfrac{{1 - {{\left( {\dfrac{1}{4}} \right)}^n}}}{{1 - \dfrac{1}{4}}}\]}
Therefore, the sum of the given series is ${\text{B}}{\text{. }}n + \dfrac{{{4^{ - n}}}}{3} - \dfrac{1}{3}$.

Note: Whenever such types of questions appear, then write down the given series and then convert into a simplified form, as mentioned in the solution i.e., \[\left( {1 - \dfrac{1}{4}} \right) + \left( {1 - \dfrac{1}{{16}}} \right) + \left( {1 - \dfrac{1}{{64}}} \right) + .......\] to n terms. Then, we know, 1+ 1+ 1+ …. to n terms is n and then the series will become \[n - \dfrac{1}{4}.\dfrac{{1 - {{\left( {\dfrac{1}{4}} \right)}^n}}}{{1 - \dfrac{1}{4}}}\], simplify it further to get the sum.