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