The product (32)(32)^ 1/6 (32)^ 1/36 ...... to , , is - Vidyadip

Question

The product $(32)(32)^{1/6}(32)^{1/36} ...... to\,\, \infty $ is

✓

Answer

c
(c) $(32)(32)^{1/6}(32)^{1/36} ......\infty $

= ${(32)^{1 + \frac{1}{6} + \frac{1}{{36}} + .....\infty }}$

$ = {(32)^{\frac{1}{{1 - (1/6)}}}}$

$ = {(32)^{\frac{1}{{5/6}}}} = {(32)^{6/5}}$

$ = {2^6} = 64$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If $f\left( x \right) = {x^3} + {e^{\frac{x}{2}}}$ and $g\left( x \right) = {f^{ - 1}}\left( x \right)$ then value of $g '(1)$ is

If $4{x^2} + p{y^2} = 45$ and ${x^2} - 4{y^2} = 5$ cut orthogonally, then the value of $p$ is

A particle moves so that $S = 6 + 48t - {t^3}$. The direction of motion reverses after moving a distance of

$\text { If } \int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} d x=A\left(\frac{\alpha x-1}{\beta x+3}\right)^B+C,$ where $\mathrm{C}$ is the constant of integration, then the value of $\alpha+\beta+20 \mathrm{AB}$ is...........

If the mean deviation about the mean of the numbers $1,2,3, \ldots ., n$, where $n$ is odd, is $\frac{5(n+1)}{n}$, then $n$ is equal to

Let $A = \left[ {\begin{array}{*{20}{c}}
p&{13}\\
{ - 13}&p
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
{4q}&{85}\\
{ - 2}&1
\end{array}} \right]$ where $p,q \in N$. It is given that $\left| A \right| = \left| B \right|$ and $p,q \in[1,1000]$. Then total number of ordered pairs $(p,q)$ is

If the function $f(x)=\left\{\begin{array}{c}\frac{\log _{e}\left(1-x+x^{2}\right)+\log _{e}\left(1+x+x^{2}\right)}{\sec x-\cos x}, x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)-\{0\} \\ k \end{array}\right.$ is continuous at $x =0$, then $k$ is equal to.

The coefficient of $x^2$ in the expansion of the product $(2 -x^2)$. $((1 + 2x + 3x^2)^6 +(1 -4x^2)^6)$ is

If $^n{P_3}{ + ^n}{C_{n - 2}} = 14n$, then $n = $

In a football championship, there were played $153$ matches. Every team played one match with each other. The number of teams participating in the championship is