Prove that : (9 1/3 . 9 1/9 . 9 1/27 ….∞) = 3.

1.	Prove that : (9 1/3 . 9 1/9 . 9 1/27 ….∞) = 3.
Answer» Using the properties of exponents: The above term can be written as Let S = \(9\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ..... \infty\) ....... (1) We observe that above progression(in power of 9) possess a common ratio. So it is a geometric progression. Let m = \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ..... \infty\) Common ratio = r = \(\cfrac{\frac{1}{9}}{\frac{1}{3}} =\frac{1}{3} \) Sum of infinite GP = \(\frac{a}{1-r}\),where a is the first term and r is the common ratio. Note: We can only use the above formula if \|r\|<1 Clearly, a = \(\frac{1}{3}\) and r = \(\frac{1}{3}\) m = \(\cfrac{\frac{1}{3}}{1-\frac{1}{3}}\) = \(\frac{1}{2}\) From equation 1 we have, S = 9^m = 9^1/2 = 3 = RHS Hence Proved

Answer»

Using the properties of exponents:

The above term can be written as

Let S = \(9\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ..... \infty\) ....... (1)

We observe that above progression(in power of 9) possess a common ratio. So it is a geometric progression.

Let m = \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ..... \infty\)

Common ratio = r = \(\cfrac{\frac{1}{9}}{\frac{1}{3}} =\frac{1}{3} \)

Sum of infinite GP = \(\frac{a}{1-r}\),where a is the first term and r is the common ratio.

Note: We can only use the above formula if |r|<1

Clearly, a = \(\frac{1}{3}\) and r = \(\frac{1}{3}\)

m = \(\cfrac{\frac{1}{3}}{1-\frac{1}{3}}\) = \(\frac{1}{2}\)

From equation 1 we have,

S = 9^m = 9^1/2 = 3 = RHS

Hence Proved

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