IPMAT Rohtak 2019 (QA) - Given (A = 2^65) and (B = 2^64 + 2^63 + 2^62 + ... + 2^0), which of the following is true? | PYQs + Solutions

IPMAT Rohtak 2019

Algebra

Progression & Series

Medium

Given $(A = 2^{65})$ and $(B = 2^{64} + 2^{63} + 2^{62} + ... + 2^0)$ , which of the following is true?

B is

(2^{64})

times larger than A

A and B are equal

B is larger than A by 1

A is larger than B by 1

Correct Option: 4

Given:

\newline

A = 2^{65}

\newline

B = 2^{64} + 2^{63} + 2^{62} + ... + 2^0

To compare A and B, let's look at the relationship between powers of 2:

\newline

B = 2^{64} + 2^{63} + 2^{62} + ... + 2^0

B

is sum of powers of 2 from

2^0

2^{64}

For a geometric series with first term

a

and ratio

r

Sum =

\frac{a(1-r^n)}{1-r}

where

n

is number of terms

Here:

\newline

a = 2^0 = 1

\newline

r = 2

\newline

n = 65

(from

2^0

2^{64}

is 65 terms)

B = \frac{1(1-2^{65})}{1-2}

B = \frac{1-2^{65}}{-1}

B = 2^{65} - 1

Therefore:

\newline

A = 2^{65}

\newline

B = 2^{65} - 1

A

is greater than

B

by 1.