Find the sum to infinity of the series
\(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\),..........
\(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\),..........
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Correct Answer: Option A
Explanation:
Sum to infinity
∑ = arn − 1
= \(\frac{a}{1}\) − r
a = \(\frac{1}{4}\)
r = \(\frac{1}{8}\) ÷ \(\frac{1}{4}\)
r = \(\frac{1}{s}\) × \(\frac{4}{1}\)
= \(\frac{1}{2}\)
S = \(\frac{1 \div 4}{1}\) − \(\frac{1}{2}\)
= \(\frac{1}{4}\) ÷ \(\frac{1}{2}\)
= \(\frac{1}{4}\) × \(\frac{2}{1}\)
= \(\frac{1}{2}\)
Sum to infinity
∑ = arn − 1
= \(\frac{a}{1}\) − r
a = \(\frac{1}{4}\)
r = \(\frac{1}{8}\) ÷ \(\frac{1}{4}\)
r = \(\frac{1}{s}\) × \(\frac{4}{1}\)
= \(\frac{1}{2}\)
S = \(\frac{1 \div 4}{1}\) − \(\frac{1}{2}\)
= \(\frac{1}{4}\) ÷ \(\frac{1}{2}\)
= \(\frac{1}{4}\) × \(\frac{2}{1}\)
= \(\frac{1}{2}\)