Question
Prove that: $\Big(9^{\frac{1}{3}}.9^{\frac19}.9^{\frac{1}{27}}\dots\infty\Big)=3.$

Answer

$9^{\frac{1}{3}}\times9^{\frac{1}{9}}\times9^{\frac{1}{27}}\dots\infty$ $=9^{\big(\frac13+\frac19+\frac{1}{27}+\ \dots\big)}$ $=9^{\Bigg(\frac{\frac13}{1-\frac13}\Bigg)}$ $\Big[\text{Using }\text{S}_\infty=\frac{\text{a}}{1-\text{r}}\Big]$ $=9^{\big(\frac{1}{3}\times\frac32\big)}$ $=9^\frac{1}{2}$ $=3$ So, $9^{\frac{1}{3}}\times9^{\frac{1}{9}}\times9^{\frac{1}{27}}\dots\infty=3$

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