Question

Given A = $ {2^{65}} $ and B = $ \left( {{2^{64}} + {2^{63}} + {2^{62}} + ... + {2^0}} \right) $
a.B is $ {2^{64}} $ larger than A
b.A and B are equal
c.B is larger than A by 1
d.A is larger than B by 1

Answer 1

Hint: In the given question, ‘B’ is in Geometric Progression that is $ {{\text{S}}_{\text{n}}} = \left( {{{\text{a}}_1} + {{\text{a}}_2} + {{\text{a}}_3} + ... + {{\text{a}}_{\text{n}}}} \right) $ where $ \dfrac{{{{\text{a}}_2}}}{{{{\text{a}}_1}}} $ gives the common ratio ‘r’. This question can be solved by solving ‘B’ using Sum to n numbers i.e., $ {{\text{S}}_{\text{n}}} = \left( {{{\text{a}}_1} + {{\text{a}}_2} + {{\text{a}}_3} + ... + {{\text{a}}_{\text{n}}}} \right) = \dfrac{{{\text{a}}\left( {{{\text{r}}^{\text{n}}} - 1} \right)}}{{{\text{r}} - 1}} $ , r $ \ne $ 1. And then based on this, we can compare B with A. We cannot solve’ A’ further so we are solving ‘B’ here.

Complete step-by-step answer:
Given, A = $ {2^{65}} $ and B = $ \left( {{2^{64}} + {2^{63}} + {2^{62}} + ... + {2^0}} \right) $
From B = $ \left( {{2^{64}} + {2^{63}} + {2^{62}} + ... + {2^0}} \right) $ , we can simplify it further by using Sum to n numbers in Geometric Progression.
Sum to n numbers, $ {{\text{S}}_{\text{n}}} = \left( {{{\text{a}}_1} + {{\text{a}}_2} + {{\text{a}}_3} + ... + {{\text{a}}_{\text{n}}}} \right) = \dfrac{{{\text{a}}\left( {{{\text{r}}^{\text{n}}} - 1} \right)}}{{{\text{r}} - 1}} $ , r $ \ne $ 1
Where, a= first term, r= common ratio, n= number of terms
 $ \Rightarrow $ B = \[\left( {{2^0} + {2^1} + {2^2} + ... + {2^{64}}} \right)\]
Here, a= $ {2^0} $ , r= $ \dfrac{{{{\text{a}}_2}}}{{{{\text{a}}_1}}} = \dfrac{{{2^1}}}{{{2^0}}} = 2 $ and n= 65
 $ \Rightarrow $ B = $ \dfrac{{{2^0}\left( {{2^{65}} - 1} \right)}}{{2 - 1}} = \dfrac{{{2^{65}} - 1}}{1} = {2^{65}} - 1 $
 $ \Rightarrow $ B = A-1
(Since A = $ {2^{65}} $ )
Therefore, A is larger than B by 1

Note: The geometric sequence is sometimes called the geometric progression or GP, for short. Here in the above question we solved the sum to n numbers using $ {{\text{S}}_{\text{n}}} = \dfrac{{{\text{a}}\left( {{{\text{r}}^{\text{n}}} - 1} \right)}}{{{\text{r}} - 1}} $ , r $ \ne $ 1. But when n= $ \infty $ , that is when the number of terms are set to infinity, then we can solve Sum to infinity using $ {{\text{S}}_\infty } = \sum\limits_{{\text{n = 1}}}^\infty {{\text{a}}{{\text{r}}^{{\text{n - 1}}}} = \dfrac{{{{\text{a}}_1}}}{{1 - {\text{r}}}}} , - 1 < r < 1 $