The value of {4^{1/3}} cdot {4^{1/9}} cdot {4 | vedclass

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

Question

(A) Let the given expression be $S = {4^{1/3}} \cdot {4^{1/9}} \cdot {4^{1/27}} \cdots \infty$.Using the law of exponents ${a^m} \cdot {a^n} = {a^{m+n

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Let the given expression be $S = {4^{1/3}} \cdot {4^{1/9}} \cdot {4^{1/27}} \cdots \infty$.Using the law of exponents ${a^m} \cdot {a^n} = {a^{m+n}}$,we get:$S = {4^{(1/3 + 1/9 + 1/27 + \cdots \infty)}}$.The exponent is an infinite geometric series with the first term $a = 1/3$ and common ratio $r = 1/3$.The sum of an infinite geometric series is given by $S_{\infty} = \frac{a}{1-r}$.Substituting the values,the sum of the exponent is $\frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = 1/2$.Therefore,$S = {4^{1/2}} = \sqrt{4} = 2$.

The value of ${4^{1/3}} \cdot {4^{1/9}} \cdot {4^{1/27}} \cdots \infty$ is

Similar Questions

Determine the $25^{th}$ term of an $A.P.$ whose $9^{th}$ term is $-6$ and the common difference is $\frac{5}{4}$.

The first term of an $A.P.$ of consecutive integers is $p^2 + 1$. The sum of $(2p + 1)$ terms of this series can be expressed as:

The value of $\sum\limits_{r = 16}^{30} {(r + 2)(r - 3)}$ is equal to

The sum of the first two terms of a $G.P.$ is $36$ and the product of the first and the third terms is $9$ times the second term,then find the sum of the first $8$ terms.

If $a_m$ denotes the $m^{th}$ term of an $A.P.$,then $a_m$ =

Vedclass Test Series

Exam Paper Generator

Online Exam Module