Given A-2-65- and B-2-64-2-63-2-62-2-0-B is 2-64- larger than AA is larger than B by 1A and B are equalB is larger than A by 1

Question

Toppr Admin · Accepted Answer

B is in G-P- with a-20-r-2-n-65-x2234-Sn-a-rn-x2212-1-r-x2212-1-20-265-x2212-1-2-x2212-1-265-x2212-1-x2234-B-265-x2212-1-x21D2-B-A-x2212-1-x2234- A is larger than B by 1-

Given $A = 2^{65}$ and $B = (2^{64} + 2^{63} + 2^{62} + . . . . + 2^{0})$
B is $2^{64}$ larger than A
A is larger than B by $1$
A and B are equal
B is larger than A by $1$

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

Given A=265 and B=(264+263+262+....+20)B is 264 larger than AA is larger than B by 1A and B are equalB is larger than A by 1

B is in G.P. with a=20,r=2,n=65∴Sn=a(rn−1)r−1=20(265−1)2−1=265−1∴B=265−1⇒B=A−1∴ A is larger than B by 1.

Given $A = 2^{65}$ and $B = (2^{64} + 2^{63} + 2^{62} + . . . . + 2^{0})$
B is $2^{64}$ larger than A
A is larger than B by $1$
A and B are equal
B is larger than A by $1$

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