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Question: The value of \(\dfrac{3}{4} + \dfrac{{15}}{{16}} + \dfrac{{63}}{{64}} + ......\) up to n terms is ...

The value of 34+1516+6364+......\dfrac{3}{4} + \dfrac{{15}}{{16}} + \dfrac{{63}}{{64}} + ...... up to n terms is
A. n−4n3−13{\text{A}}{\text{. }}n - \dfrac{{{4^n}}}{3} - \dfrac{1}{3}
B. n+4−n3−13{\text{B}}{\text{. }}n + \dfrac{{{4^{ - n}}}}{3} - \dfrac{1}{3}
C. n+4n3−13{\text{C}}{\text{. }}n + \dfrac{{{4^n}}}{3} - \dfrac{1}{3}
D. n−4−n3+13{\text{D}}{\text{. }}n - \dfrac{{{4^{ - n}}}}{3} + \dfrac{1}{3}

Explanation

Solution

Hint: Write the series 34+1516+6364+......\dfrac{3}{4} + \dfrac{{15}}{{16}} + \dfrac{{63}}{{64}} + ...... as (1−14)+(1−116)+(1−164)+.......\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 34+1516+6364+......\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-
(1−14)+(1−116)+(1−164)+.......\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.