Math, asked by rishabhdhakad20000, 3 months ago

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

Answers

Answered by anindyaadhikari13
6

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)

Answered by Anisha5119
4

Answer:

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

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