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

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

Question

(A) The given expression is $P = 2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \dots \infty$. Expressin

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given expression is $P = 2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \dots \infty$. Expressing all terms with base $2$: $P = 2^{\frac{1}{4}} \cdot (2^2)^{\frac{1}{16}} \cdot (2^3)^{\frac{1}{48}} \cdot (2^4)^{\frac{1}{128}} \cdot \dots$ $P = 2^{\frac{1}{4}} \cdot 2^{\frac{2}{16}} \cdot 2^{\frac{3}{48}} \cdot 2^{\frac{4}{128}} \cdot \dots$ $P = 2^{\frac{1}{4}} \cdot 2^{\frac{1}{8}} \cdot 2^{\frac{1}{16}} \cdot 2^{\frac{1}{32}} \cdot \dots$ Using the property $a^m \cdot a^n = a^{m+n}$,we get: $P = 2^{(\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \dots)}$ The exponent is an infinite geometric series with first term $a = \frac{1}{4}$ and common ratio $r = \frac{1}{2}$. The sum of an infinite geometric series is $S = \frac{a}{1-r}$. $S = \frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{2}$. Therefore,$P = 2^{\frac{1}{2}}$.

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

Similar Questions

The $4^{\text{th}}$ term of a $GP$ is $500$ and its common ratio is $\frac{1}{m}$,where $m \in N$. Let $S_n$ denote the sum of the first $n$ terms of this $GP$. If $S_6 > S_5+1$ and $S_7 < S_6+\frac{1}{2}$,then the number of possible values of $m$ is $..........$

If the arithmetic mean and geometric mean of the $p^{\text{th}}$ and $q^{\text{th}}$ terms of the sequence $-16, 8, -4, 2, \ldots$ satisfy the equation $4x^{2}-9x+5=0$,then $p+q$ is equal to ..... .

What will $Rs. 500$ amount to in $10$ years after its deposit in a bank which pays an annual interest rate of $10\%$ compounded annually?

If $\alpha, \beta$ are the roots of the equation $x^2 - 3x + a = 0$ and $\gamma, \delta$ are the roots of the equation $x^2 - 12x + b = 0$,and if $\alpha, \beta, \gamma, \delta$ form an increasing $G.P.$,then $(a, b) = $

If the ratio of the sum of the first three terms and the sum of the first six terms of a $G.P.$ is $125 : 152$,then the common ratio $r$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module