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Prove that:`3^(1/2)xx3^(1/4)xx3^(1/8)xx...=3`

Answer» `L.H.S. = 3^(1/2)xx3^(1/4)xx3^(1/8)+...`
`=3^(1/2+1/4+1/8+1/16...)`
Now, we will solve, `(1/2+1/4+1/8+1/16...)`
It forms a GP with first term, `a= 1/2` and common ratio, `r = 1/2`
So, Sum of the given series will be, `S = a/(1-r) = (1/2)/(1-1/2) = 1`
So, our expression becomes,
`3^(1/2+1/4+1/8...) = 3^1 = 3 = R.H.S.`


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