Question

prove that 6^1/2 ×6^1/9 ×6^1/27 ×..... ......00 = √6​

Answer 1

Required Answer:-

Given to prove:

$\rm \mapsto {6}^{ \frac{1}{3}} \times {6}^{ \frac{1}{9} } \times {6}^{ \frac{1}{27} } .... \infty = \sqrt{6}$

Proof:

Taking LHS,

$\rm {6}^{ \frac{1}{3}} \times {6}^{ \frac{1}{9} } \times {6}^{ \frac{1}{27} } .... \infty$

$\rm = {6}^{ \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ... \infty }$

Let us assume that,

$\rm S = \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + ..... \dfrac{1}{ {3}^{n} }$

$\rm \implies 3S = 1 + \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + ..... \dfrac{1}{ {3}^{n} }$

$\rm \implies 3S - S = 1$

$\rm \implies 2S= 1$

$\rm \implies S= \dfrac{1}{2}$

Hence, the sum of the series is 1/2.

So,

$\rm {6}^{S}$

$\rm = {6}^{ \dfrac{1}{2} }$

$\rm = \sqrt{6}$

= RHS (Hence Proved)

Answer 2

Answer:

LHS=RHS HENCE PROVED...........