Question

The value of $9^{\frac{1}{3}}{.9}^{\frac{1}{9}}{.9}^{\frac{1}{27}}...........$ upto $\infty$, is

• A. $1$
• B. $3$
• C. $9$
• D. $Noneofthese$

Answer 1

Answer: (B)

$3$

$9^{\frac{1}{3}}{.9}^{\frac{1}{9}}{.9}^{\frac{1}{27}}----upto\infty$

$=9^{\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+---\right)}$

$=9^{\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)}$

$=9^{\left(\frac{\frac{1}{3}}{\frac{2}{3}}\right)}$

$=9^{\frac{1}{2}}$

$=3$