Question

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

• 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

The correct option is D A is larger than B by 1

$BisinG.Pwitha=2^{\circ },r=2,n=65S_n=\frac{\left[a\right(r^n-1\left)\right]}{(r-1)}=\frac{\left[20\right(2^{65}–1\left)\right]}{(2-1)}B=2^{65}-1$
B = A-1
A is larger than B by 1.