Question : What is the simplified value of $(3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)$?
Option 1: $\frac{(3^{32}-1)}{2}$
Option 2: $\frac{(3^{16}-1)}{2}$
Option 3: $\frac{(3^{64}-1)}{2}$
Option 4: $\frac{(3^{128}-1)}{2}$
Correct Answer: $\frac{(3^{32}-1)}{2}$
Solution :
Given: $(3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)$
Multiplying and dividing by 2, we get:
$\frac{1}{2}(3-1)(3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)$
= $\frac{1}{2}(3^2-1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)$
= $\frac{1}{2}(3^4-1)(3^{4}+1)(3^{8}+1)(3^{16}+1)$
= $\frac{1}{2}(3^8-1)(3^{8}+1)(3^{16}+1)$
= $\frac{1}{2}(3^{16}-1)(3^{16}+1)$
= $\frac{1}{2}(3^{32}-1)$
= $\frac{3^{32}-1}{2}$
Hence, the correct answer is $\frac{3^{32}-1}{2}$.
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