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

Given:

$A = 2^{65}$

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

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

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

$B$ is sum of powers of 2 from $2^0$ to $2^{64}$

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:

$a = 2^0 = 1$

$r = 2$

$n = 65$ (from $2^0$ to $2^{64}$ is 65 terms)

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

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

$B = 2^{65} - 1$

Therefore:

$A = 2^{65}$

$B = 2^{65} - 1$

$A$ is greater than $B$ by 1.