Question

If $ {s_n} = \sum\nolimits_{n = 1}^n {\dfrac{{1 + 2 + {2^2} + ...to\,n\,terms}}{{{2^n}}}} $ then $ {s_n} $ is equal to
A. $ {2^n} - (n + 1) $
B. $ 1 - \dfrac{1}{{{2^n}}} $
C. $ n - 1 + \dfrac{1}{{{2^n}}} $
D. $ {2^n} - 1 $

Answer 1

Hint: In this question, we are given an expression that tells us the value of $ {s_n} $ , $ {s_n} $ means the sum of all the terms up to n terms and $ \sum {} $ (sigma) is used to show the sum of all the terms from one value of the variable written inside it to the other value. In this question, we have to find the sum of the expression written in it at values 1, 2 up to n. The expression written as a fraction, a fraction is divided into two parts namely numerator and denominator by a horizontal line. The numerator of the given fraction is a geometric progression. So, we will first simplify the given fraction by using the formula of the sum of n terms of a G.P. and then canceling out the common terms if present.

Complete step-by-step answer:
We are given $ {s_n} = \sum\nolimits_{n = 1}^n {\dfrac{{1 + 2 + {2^2} + ...to\,n\,terms}}{{{2^n}}}} $
We know that for a G.P. $ {s_n} = \dfrac{{a({r^n} - 1)}}{{r - 1}} $
 $ 1 + 2 + {2^2}{....2^n} $ is a G.P.
So, $ 1 + 2 + {2^2} + {...2^n} = \dfrac{{1({2^n} - 1)}}{{2 - 1}} = {2^n} - 1 $
 $
   \Rightarrow {s_n} = \sum\nolimits_{n = 1}^n {\dfrac{{{2^n} - 1}}{{{2^n}}}} \\
   \Rightarrow {s_n} = \sum\nolimits_{n = 1}^n {1 - \dfrac{1}{{{2^n}}}} \;
  $
We know that $ \sum\limits_{n = 1}^n 1 = n $
 $ \Rightarrow {s_n} = n - \sum\nolimits_{n = 1}^n {\dfrac{1}{{{2^n}}}} $
Now, $ \sum\nolimits_{n = 1}^n {\dfrac{1}{{{2^n}}}} $ is a G.P. whose first term is $ \dfrac{1}{2} $ and the common ratio is $ \dfrac{1}{2} $ .
So,
 $
  \sum\nolimits_{n = 1}^n {\dfrac{1}{{{2^n}}}} = \dfrac{{\dfrac{1}{2}[{{(\dfrac{1}{2})}^n} - 1]}}{{\dfrac{1}{2} - 1}} \\
   \Rightarrow \sum\nolimits_{n = 1}^n {\dfrac{1}{{{2^n}}}} = \dfrac{{\dfrac{1}{2}(\dfrac{1}{{{2^n}}} - 1)}}{{\dfrac{{ - 1}}{2}}} \\
   \Rightarrow \sum\nolimits_{n = 1}^n {\dfrac{1}{{{2^n}}}} = 1 - \dfrac{1}{{{2^n}}} \\
  $
On putting this value in the obtained expression for $ {s_n} $ , we get –
 $
  {s_n} = n - (1 - \dfrac{1}{{{2^n}}}) \\
   \Rightarrow {s_n} = n - 1 + \dfrac{1}{{{2^n}}} \;
  $
Hence, option (C) is the correct answer.
So, the correct answer is “Option C”.

Note: A geometric progression is defined as a series or progression in which the ratio of any two consecutive terms of the sequence is constant, this constant value is known as the common ratio of the G.P. A geometric progression is of the form $ a,ar,a{r^2}.... $ . In this question, we obtained two G.P. and we found their sum as we knew their first term and common ratio.