Correct Answer - Option 4 :
\(\frac{4}{3}\)
Concept:
For a GP of the form:
a + ar + ar2 + ar3 +…..
The sum is given by:
\(Sum= \frac{a}{{1 - r}}\)
Calculation:
The given sequence is
\(x\left( n \right) = 1 + \frac{1}{4} + \frac{1}{{16}} + \frac{1}{{64}} + \frac{1}{{256}} + \ldots\)
\(= 1 + \frac{1}{{{2^2}}} + \frac{1}{{{2^4}}} + \frac{1}{{{2^6}}} + \frac{1}{{{2^8}}} + \ldots \)
The series is GP series with:
a = 1
\(r = \frac{1}{{{2^2}}} = \frac{1}{4}\)
Sum of the infinite series will be:
\(Sum= \frac{a}{{1 - r}}\)
\(= \frac{1}{{1 - \frac{1}{4}}} = \frac{1}{{\frac{3}{4}}}\)
\(= \frac{4}{3}\)
