Question

The sum of n terms of the series \[\dfrac{1}{2} + \dfrac{3}{4} + \dfrac{7}{8} + \dfrac{{15}}{{16}} + \ldots \] is
A. \[{2^{ - n}}\]
B. \[{2^{ - n}}(n - 1)\]
C. \[{2^n}(n - 1) + 1\]
D. \[{2^{ - n}} + n - 1\]

Answer 1

Hint: We know that to solve the given series first rewrite the numerator of the fraction series as 1 subtracted from the value of the denominator and then simplify the equation to get the desired result.

Formula Used: \[\sum\limits_1^n 1 = n\] and \[\sum\limits_{k = 1}^n {\dfrac{1}{{{2^k}}}} = 1 - \dfrac{1}{{{2^n}}}\].

Complete step-by-step solution:
We are given a fraction series that is \[\dfrac{1}{2} + \dfrac{3}{4} + \dfrac{7}{8} + \dfrac{{15}}{{16}} + \ldots \]
Now we rewrite the numerator of the fraction series and then simplify the equation, we get
\[
  S = \dfrac{{(2 - 1)}}{2} + \dfrac{{(4 - 1)}}{4} + \dfrac{{(8 - 1)}}{8} + \dfrac{{(16 - 1)}}{{16}} + \ldots .n \\
   = \left( {\dfrac{2}{2} - \dfrac{1}{2}} \right) + \left( {\dfrac{4}{4} - \dfrac{1}{4}} \right) + \left( {\dfrac{8}{8} - \dfrac{1}{8}} \right) + \left( {\dfrac{{16}}{{16}} - \dfrac{1}{{16}}} \right) + \ldots .n \\
   = \left( {1 + 1 + 1 \ldots n} \right) - \left( {\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \ldots .n} \right) \\
   = \left( {\sum\limits_1^n 1 } \right) - \left( {\sum\limits_{k = 1}^n {\dfrac{1}{{{2^k}}}} } \right)
 \]
Further Simplifying, we get,
\[
  S = {\text{n}} - \left( {1 - \dfrac{1}{{{2^{\text{n}}}}}} \right) \\
   = n - 1 + {2^{ - n}}
 \]
Therefore, the sum of \[n\] terms of the series \[\dfrac{1}{2} + \dfrac{3}{4} + \dfrac{7}{8} + \dfrac{{15}}{{16}} + \ldots \] is \[{2^{ - n}} + n - 1\].

So, option D is correct

Additional information: When a series of numbers repeats itself in a sequential manner, it is then known as a sequential series. Arithmetic is based on the concepts of sequence and series. A series is the total of all components, but a sequence is an itemized collection of items that allow for any sort of recurrence. In the solution as 1 is been added n times, so it can be written as the product of 1 and n which is n, that gives the same result. While solving the series it is better to write a term in numerator same as that of the denominator, so it may give arithmetic or a geometric progression which are easier to solve by formula.

Note Many students make miscalculations while writing the values of numerator and denominator so make sure about the values of the fraction.
We can also solve the summation using the GP formula as well which is Sum = $\dfrac{a(r^n-1)}{r-1}$ where r is not equal to 1. Using this we will find the sum of the second part of the series.
For the GP: $ \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \ldots .n$,
a = $\dfrac{1}{2}, r =\dfrac{1}{2}$
S = $\dfrac{\dfrac{1}{2}(\dfrac{1}{2}^n-1)}{\dfrac{1}{2}-1}$
S = $1-\dfrac{1}{2}^n$