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The value of the product \({6^{\frac{1}{2}}} \times {6^{\frac{1}{4}}} \times {6^{\frac{1}{8}}} \times {6^{\frac{1}{{16}}}} \times \ldots \) up to infinite terms is
1. 6
2. 36
3. 216
4. 512
5. None of these

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Correct Answer - Option 1 : 6

CONCEPT:

Infinite Geometric series:

Standard infinite geometric series is given by: a1 + a1r + a1r2 + a1r3 + …… + a1rn-1 + ….

Sum of infinite geometric series is given by \({{\rm{s}}_{\rm{n}}} = \frac{{\rm{a}}}{{1 - {\rm{r}}}}\) as long as -1 < r < 1

CALCULATION:

Given that: \({6^{\frac{1}{2}}} \times {6^{\frac{1}{4}}} \times {6^{\frac{1}{8}}} \times {6^{\frac{1}{{16}}}} \times \cdots\) upto infinite terms

\({6^{\left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots } \right)}}\)     ...(∵ am × an = am+n)

The series: \(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots\)is in form of infinite geometric series.

Comparing it with standard infinite G.P series, we get a = 1/2 and r = 1/2

Sum of infinite terms of a G.P, \({s_n} = \frac{a}{{1 - r}}\)

\( \Rightarrow {{\rm{s}}_{\rm{n}}} = \left( {\frac{{\frac{1}{2}}}{{1 - \frac{1}{2}}}} \right)\)

\(\Rightarrow {{\rm{s}}_{\rm{n}}} = \left( {\frac{{\frac{1}{2}}}{{\frac{1}{2}}}} \right) = 1\)

\(\Rightarrow {6^{\left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots } \right)}} = {6^1} = 6\)

Hence, the value of product of \({6^{\frac{1}{2}}} \times {6^{\frac{1}{4}}} \times {6^{\frac{1}{8}}} \times {6^{\frac{1}{{16}}}} \times \cdots\) upto infinite terms = 6

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