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

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: Option D
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Solution(By Apex Team)

B is in G.P. with a = $2^{0}$, r = 2, n = 65 $\begin{array}{l}\begin{aligned}\therefore S_n&=\frac{a\left(r^n-1\right)}{r-1}\\ &=\frac{2^0\left(2^{65}-1\right)}{2-1}\\ \therefore B&=2^{65}-1\\ \Rightarrow B&=A-1\end{aligned}\\ \therefore A\text{ is larger than }B\text{ by }1\end{array}$

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