Question

Sum of the series $9^{\frac{1}{3}}\times 9^{\frac{1}{9}}\times 9^{\frac{1}{27}}\times .......$ is equal to

• A. $3$
• B. $9$
• C. $27$
• D. $81$

Answer 1

The correct option is A $3$

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

$=9^{\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...}$

Let $S=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...$

$=\frac{\frac{1}{3}}{1-\frac{1}{3}}$

$=\frac{\frac{1}{3}}{\frac{2}{3}}=\frac{1}{2}$

$\therefore 9^{\frac{1}{3}}\times 9^{\frac{1}{9}}\times 9^{\frac{1}{27}}\times ...\infty =9^{\frac{1}{2}}$

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

$=3$