Question

Given $A=2^{65}$ and $B=(2^{64}+2^{63}+2^{62}+....+2^0)$

• 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

Answer: (D)

A is larger than B by $1$

B is in G.P. with $a=2^0,r=2,n=65$
$\therefore S_n=\frac{a(r^n-1)}{r-1}=\frac{2^0(2^{65}-1)}{2-1}=2^{65}-1$
$\therefore B=2^{65}-1$
$\Rightarrow B=A-1$
$\therefore$ A is larger than B by $1$.