Given: \(S=3+2+\frac{4}{3}+\frac{8}{9}+...\)

We know that sum(S) of infinite geometric series is given by

\(S=\frac{a}{1-r}\)

Where a is first term and r is common ratio

Here,

\(a=3,r=\frac{2}{3}\)

So, by using equation(1) sum(S) of infinite geometric series will be

\(S=\frac{3}{1-\frac{2}{3}}\)

\(S=\frac{3}{(\frac{3-2}{3})}\)

\(S=\frac{3}{(\frac{1}{3})}\)

\(=3(3)\)

\(=9\)

Hence, sum of given infinite geometric series is 9.

We know that sum(S) of infinite geometric series is given by

\(S=\frac{a}{1-r}\)

Where a is first term and r is common ratio

Here,

\(a=3,r=\frac{2}{3}\)

So, by using equation(1) sum(S) of infinite geometric series will be

\(S=\frac{3}{1-\frac{2}{3}}\)

\(S=\frac{3}{(\frac{3-2}{3})}\)

\(S=\frac{3}{(\frac{1}{3})}\)

\(=3(3)\)

\(=9\)

Hence, sum of given infinite geometric series is 9.