The value of 9^1/3 . 9^1/9 . 9^1/27 ... upto inf, is
(a) 1
(b) 3
(c) 9
(d) none of these

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Solution

$(b) 3 9^{\frac{1}{3}} \times 9^{\frac{1}{9}} \times 9^{\frac{1}{27}} \times . . . \infty = 9^{(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + . . . \infty)} Here, it is a G . P . with a = \frac{1}{3} and r = \frac{1}{3} . ∴ 9^{(\frac{\frac{1}{3}}{1 - \frac{1}{3}})} = 9^{(\frac{1}{2})} = 3$

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Similar questions

9^{1 / 3} \times 9^{1 / 9} \times 9^{1 / 27} \times . . . . . \infty

The value of $9^{\frac{1}{3}} {.9}^{\frac{1}{9}} {.9}^{\frac{1}{27}} . . . . .$ to $\infty$ is

9^{\frac{1}{3}} {.9}^{\frac{1}{9}} {.9}^{\frac{1}{27}} . . . . . . . . . . .

\infty

Prove that :

$(9^{\frac{1}{3}} {.9}^{\frac{1}{9}} {.99}^{\frac{1}{27}} {.169}^{\frac{1}{32}} . . . . . \infty) = 3.$

x = 9^{1 / 3} \cdot 9^{1 / 9} \cdot 9^{1 / 27} \dots \infty; y = 4^{1 / 3} \cdot 4^{- 1 / 9} \cdot 4^{1 / 27} \dots \infty

z = \infty \sum r = 1 (1 + i)^{- r}

p = x + y z

- {tan}^{- 1} \frac{\sqrt{a}}{b}

a^{2} + b^{2}

a

b

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