Find the sum to infinity of the series
\(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\),..........

A.

\(\frac{1}{2}\)

B.

\(\frac{3}{5}\)

C.

\(\frac{-1}{5}\)

D.

\(\frac{73}{12}\)

Correct answer is A

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 = \(\frac{1 \div 4}{1}\) − \(\frac{1}{2}\)

= \(\frac{1}{4}\) ÷ \(\frac{1}{2}\)

= \(\frac{1}{4}\) × \(\frac{2}{1}\)

= \(\frac{1}{2}\)