Question

$A=(2+1)(2^2+1)(2^4+1).....\left(2^{2016}+1\right)$. The value of ${\left(A+1\right)}^{\frac{1}{2016}}$ is

• A. $4$
• B. $2016$
• C. ${}^{2^{4032}}$
• D. $2$

Answer 1

The correct option is A $4$

$A=(2+1)(2^2+1)(2^4+1)(2^8+1).....(2^{2016}+1)$

$A=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)....(2^{2016}+1)$

$A=(2^2-1)(2^2+1)(2^4+1)......(2^{2016}+1)$

$A=(2^4-1)(2^4+1)......(2^{2016}+1)$

Further simplification gives

$A=(2^{2016}-1)(2^{2016}+1)$

$A=2^{4032}-1$

$\Rightarrow A+1=2^{4032}$

$\Rightarrow (A+1{)}^{1/2016}=(2^{4032}{)}^{1/2016}$

$=2^2$

$=4$