Question

If $P=3^{\frac{1}{3}}.3^{\frac{2}{9}}.3^{\frac{3}{27}}....\infty$, then $P^{\frac{4}{3}}=$

• A. $3^{\frac{1}{4}}$
• B. $3$
• C. $9$
• D. $3^{\frac{3}{4}}$

Answer 1

Answer: (B)

$3$

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

$P=3^{\frac{1}{3}+\frac{2}{9}+\frac{3}{27}+....+\infty }$

Let $P=3^S$

$S=\frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\frac{4}{81}+......$ ....(1)

$S=\frac{1}{3}[1+\frac{2}{3}+\frac{3}{9}+.....\infty ]$

$3S=1+\frac{2}{3}+\frac{3}{9}+\frac{4}{27}+.....\infty$ ....(2)

Subtracting (1) from (2) we get,

$2S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+.....\infty$

or, $2S=1+\frac{1}{3}\left(\frac{1}{1-\frac{1}{3}}\right)$

or, $2S=1+\frac{1}{3}\times \frac{3}{2}$

or, $2S=\frac{3}{2}$

or, $S=\frac{3}{4}$

So, $P=3^{\frac{3}{4}}$

or, $P^{\frac{4}{3}}=[3^{\frac{3}{4}}{]}^{\frac{4}{3}}$

or, $P^{\frac{4}{3}}=3$