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: $2^{64}-1$
Correct Answer: $2^{16}-1$
Solution : Given: $(2+1)(2^{2}+1)(2^{4}+1)(2^{8}+1)$ Multiplying (2 – 1) = 1 in the equation, we get: $(2-1)(2+1)(2^{2}+1)(2^{4}+1)(2^{8}+1)$ = $(2^2-1)(2^{2}+1)(2^{4}+1)(2^{8}+1)$ = $(2^4-1)(2^{4}+1)(2^{8}+1)$ = $(2^8-1)(2^{8}+1)$ = $(2^{16}-1)$ Hence, the correct answer is $(2^{16}-1)$.
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Question : What is the simplified value of $(3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)$?
Question : What is the simplified value of $(x^{32}+\frac{1}{x^{32}})(x^{8}+\frac{1}{x^{8}})(x-\frac{1}{x})(x^{16}+\frac{1}{x^{16}})(x+\frac{1}{x})(x^{4}+\frac{1}{x^{4}})?$
Question : What is the simplified value of $(x^{128}+1)(x^{32}+1)(x^{64}+1)(x^{16}+1)(x^{8}+1)(x^{4}+1)(x^{2}+1)(x+1)?$
Question : If $2^{2x-y}=16$ and $2^{x+y}=32$, the value of $xy$ is:
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 $2 \frac{1}{4}-\frac{3+4}{3^3+4^3}$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile