Find the sum to infinity of the series
\(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\),……….
-
\(\frac{1}{2}\)
- \(\frac{3}{5}\)
- \(\frac{-1}{5}\)
- \(\frac{73}{12}\)
The correct answer is: 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 = \(\frac{1 \div 4}{1}\) − \(\frac{1}{2}\)
= \(\frac{1}{4}\) ÷ \(\frac{1}{2}\)
= \(\frac{1}{4}\) × \(\frac{2}{1}\)
= \(\frac{1}{2}\)
There is an explanation video available .