Question : What is the value of (22 + 1)(24+1)(28 + 1)(216 + 1) .........(2128 + 1)?
Option 1: $\frac{2^{256}-1}{2}$
Option 2: $\frac{2^{256}-1}{3}$
Option 3: $2^{256}-1$
Option 4: $\frac{2^{256}-1}{4}$
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Correct Answer: $\frac{2^{256}-1}{3}$
Solution :
$(2^2 + 1) (2^4 + 1) (2^8 + 1) (2^{16} + 1) ........ (2^{128} + 1)$
Divide by $(2^2 - 1)$ in numerator and denominator
= $\frac{1}{(2^2 - 1)}[(2^2 - 1)(2^2 + 1) (2^4 + 1) (2^8 + 1) (2^{16} + 1) ........ (2^{128} + 1)]$
= $\frac{1}{(2^2 - 1)}[(2^4 - 1) (2^4 + 1) (2^8 + 1) (2^{16} + 1) ........ (2^{128} + 1)]$
= $\frac{1}{(2^2 - 1)}[(2^8 - 1) (2^8 + 1) (2^{16} + 1) ........ (2^{128} + 1)]$
= $\frac{1}{(2^2 - 1)}[(2^{16} - 1) (2^{16} + 1) ........ (2^{128} + 1)]$
= $\frac{1}{(2^2 - 1)}(2^{256} - 1)$
= $\frac{(2^{256} - 1)}{3}$
Hence, the correct answer is $\frac{(2^{256} - 1)}{3}$.
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