ABC26GN1967 · Surds and Indices
Subject: General Aptitude · Chapter: Surds and Indices · Exam: 2005 · Marks: · Difficulty:
$1+(3+1)\left(3^{2}+1\right)\left(3^{4}+1\right)\left(3^{8}+1\right)\left(3^{16}+1\right)$ $\left(3^{32}+1\right)$ is equal to
(a)$\frac{3^{64}-1}{2}$
(b)$\frac{3^{64}+1}{2}$
(c)$3^{64}-1$
(d)$3^{64}+1$
Answer
Explanation
$1+(3+1)\left(3^{2}+1\right)\left(3^{4}+1\right)\left(3^{8}+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)$ $$\begin{aligned} = & 1+\frac{1}{2}\left[(3-1)(3+1)\left(3^{2}+1\right)\left(3^{4}+1\right)\left(3^{8}+1\right)\right. \\ & \left.\left(3^{16}+1\right)\left(3^{32}+1\right)\right] \\ = & 1+\frac{1}{2}\left[\left(3^{2}-1\right)\left(3^{2}+1\right)\left(3^{4}+1\right)\left(3^{8}+1\right)\right. \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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