JEE Main 2020 (Online) 9th January Morning ShiftQuestion 1
The product $2^{\frac14}\cdot4^{\frac1{16}}\cdot8^{\frac1{48}}\cdot16^{\frac1{128}}\cdots$ to $\infty$ is equal to :
Sequence and Series · Geometric Progression
Solution
Formulas used. (1) $2^{p}\cdot2^{q}\cdots=2^{p+q+\cdots}$(2) Sum of an infinite G.P.: $\frac{\text{first}}{1-r}$Step 1 - Write every factor as a power of $2$. $$2^{\frac14},\quad 4^{\frac1{16}}=2^{\frac2{16}}=2^{\frac18},\quad 8^{\frac1{48}}=2^{\frac3{48}}=2^{\frac1{16}},\quad 16^{\frac1{128}}=2^{\frac4{128}}=2^{\frac1{32}},\ \dots$$ Step 2 - Add the exponents. $$\frac14+\frac18+\frac1{16}+\cdots=\frac{1/4}{1-1/2}=\frac12 .$$ Step 3. The product is $2^{\frac12}$. Answer: (B)