Question:
Prove that $9^{1 / 3} \times 9^{1 / 9} \times 9^{1 / 27} \times \ldots \ldots^{\infty}=3$
Solution:
L.H.S $=9^{1 / 3} \times 9^{1 / 9} \times 9^{1 / 27} \times \ldots \ldots \infty$
$=9^{(1 / 3)+(1 / 9)+(1 / 27)+\ldots \infty}$
The series in the exponent is an infinite geometric series
Whose, $a=\frac{1}{3}$
$r=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{1 \times 3}{1 \times 9}=\frac{1}{3}$
∴Sum of the series in the exponent $=\frac{a}{1-r}=\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1 \times 3}{3 \times 2}=\frac{1}{2}$
$\therefore \mathrm{L} . \mathrm{H} . \mathrm{S}=9^{1 / 2}$
=3=R.H.S
Hence, Proved that $9^{1 / 3} \times 9^{1 / 9} \times 9^{1 / 27} \times \ldots \ldots \infty=3$