Prove that

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 $9^{1 / 3} \times 9^{1 / 9} \times 9^{1 / 27} \times \ldots \ldots \infty=3$

 

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