The product 2^{frac{1}{4}} cdot 4^{frac{1}{16 | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \ldots . \infty$

$=2^{\frac{1}{4}} \cdot 2^{\frac{2}{

Vedclass · Accepted Answer

$2^{\frac{1}{2}}$

Vedclass · Answer

$2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \ldots . \infty$

$=2^{\frac{1}{4}} \cdot 2^{\frac{2}{16}} \cdot 2^{\frac{3}{48}} \cdot 2^{\frac{4}{128}} \cdot \ldots \infty$

$=2^{\frac{1}{4}} \cdot 2^{\frac{1}{8}} \cdot 2^{\frac{1}{16}} \cdot 2^{\frac{1}{32}} \cdot \ldots \infty$

$=2^{\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\ldots \infty}{32}=(2)\left(\frac{1 / 4}{1-1 / 2}\right)=2^{1 / 2}$

The product $2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \ldots .$ to $\infty$ is equal to

Similar Questions

$0.14189189189….$ can be expressed as a rational number

If $a,\;b,\;c$ are in $A.P.$, then ${3^a},\;{3^b},\;{3^c}$ shall be in

Find the sum of the products of the corresponding terms of the sequences $2,4,8,16,32$ and $128,32,8,2, \frac{1}{2}$

Which term of the following sequences:

$\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ is $\frac{1}{19683} ?$

If the first and the $n^{\text {th }}$ term of a $G.P.$ are $a$ and $b$, respectively, and if $P$ is the product of $n$ terms, prove that $P^{2}=(a b)^{n}$