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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Question : What is the simplified value of $(x^{32}+\frac{1}{x^{32}})(x^{8}+\frac{1}{x^{8}})(x-\frac{1}{x})(x^{16}+\frac{1}{x^{16}})(x+\frac{1}{x})(x^{4}+\frac{1}{x^{4}})?$
Option 1: $(x^{64}+\frac{1}{x^{64}})$
Option 2: $\frac{(x^{64}-\frac{1}{x^{64}})}{(x^2+\frac{1}{x^2})}$
Option 3: $\frac{(x^{64}-\frac{1}{x^{64}})}{(x+\frac{1}{x})}$
Option 4: $\frac{(x^{32}-\frac{1}{x^{32}})}{(x+\frac{1}{x})}$
Question : What is the simplified value of $(x^{128}+1)(x^{32}+1)(x^{64}+1)(x^{16}+1)(x^{8}+1)(x^{4}+1)(x^{2}+1)(x+1)?$
Option 1: $x^{256}-1$
Option 2: $\frac{x^{128}-1}{x-1}$
Option 3: $\frac{x^{64}-1}{x-1}$
Option 4: $\frac{x^{256}-1}{x-1}$
Question : What is the simplified value of $(2+1)(2^{2}+1)(2^{4}+1)(2^{8}+1)$?
Option 1: $2^{8}-1$
Option 2: $2^{16}-1$
Option 3: $2^{32}-1$
Option 4: $2^{64}-1$
Question : What is the simplified value of $\left(1-\frac{1}{4-\frac{2}{1+\frac{1}{\frac{1}{3}+2}}}\right) \times \frac{15}{16} \div \frac{2}{3}$ of $2 \frac{1}{4}-\frac{3+4}{3^3+4^3}$
Option 1: $\frac{5}{13}$
Option 2: $\frac{4}{13}$
Option 3: $\frac{8}{13}$
Option 4: $\frac{6}{13}$
Question : What is the simplified value of: $7 \frac{1}{3} \div 2 \frac{1}{2}$ of $1 \frac{3}{5}-\left(\frac{3}{8}+\frac{1}{7} \times 1 \frac{3}{4}\right)-\frac{5}{24}$
Option 1: $1$
Option 2: $2$
Option 3: $\frac{1}{24}$
Option 4: $\frac{1}{12}$
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