Consider the given series as

\(I=\sum_{n=0}^\infty(\frac{5}{2^n}+\frac{1}{3^n})\)

To find the few terms ,substitute the value of n from 0 to infinity in given series i.e.

\(I=(\frac{5}{2^0}+\frac{1}{3^0})+(\frac{5}{2^1}+\frac{1}{3^1})+(\frac{5}{2^2}+\frac{1}{3^2})+...\)

\(=(5+1)+(\frac{15+2}{6})+(\frac{45+4}{36})+...\)

\(=6+\frac{17}{6}+\frac{49}{36}+...\)

Since the given series is geometric series. So, use formula of geometric sum of infinite series

\(S_n=\frac{a(1-r^n)}{1-r}\)

where a is the first term of series and r is geometric ratio between every two terms.

As, the given series can be written as

\(I=\sum_{n=0}^\infty(\frac{5}{2^n})+\sum_{n=0}^\infty(\frac{1}{3^n})\)

If the series

\(\sum a_n=\sum_{n=0}^\infty(\frac{5}{2^n})\quad\text{and}\quad\sum b_n=\sum_{n=0}^\infty\frac{1}{3^n}\)

are convergent series then it can be written in the form of

\(\sum a_n+b_n=\sum a_n+\sum b_n\)

Check the convergence of the series by ratio test

\(\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow\infty}\frac{\frac{5}{2^{n+1}}}{\frac{5}{2^n}}\)

\(=\lim_{n\rightarrow\infty}\frac{2^n}{2^{n+1}}\)

\(=\lim_{n\rightarrow\infty}\frac{2^n}{2^n(2)}\)

\(=\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\frac{1}{2}<1\)</span>

As, the limit for above series is less than 1 . Therefore, the above series is convergent.

Now, check for another series

\(=\lim_{n\rightarrow\infty}\frac{b_{n+1}}{b_n}=\frac{\frac{1}{3^{n+1}}}{\frac{1}{3^n}}\)

\(=\lim_{n\rightarrow\infty}\frac{3^n}{3^n(3)}\)

\(=\lim_{n\rightarrow\infty}\frac{b_{n+1}}{b_n}=\frac{1}{3}<1\)</span>

It implies that the limit of above series is less than 1 and hence is convergent.

Therefore, the given series is

\(I=5\sum_{n=0}^\infty(\frac{1}{2^n})+\sum_{n=0}^{\infty}(\frac{1}{3^n})\)

By using formula of sum of infinite series

\(5(2)+\frac{3}{2}\)

\(\frac{23}{2}\)

Hence,the sum of given infinite series is \(\frac{23}{2}\)

\(I=\sum_{n=0}^\infty(\frac{5}{2^n}+\frac{1}{3^n})\)

To find the few terms ,substitute the value of n from 0 to infinity in given series i.e.

\(I=(\frac{5}{2^0}+\frac{1}{3^0})+(\frac{5}{2^1}+\frac{1}{3^1})+(\frac{5}{2^2}+\frac{1}{3^2})+...\)

\(=(5+1)+(\frac{15+2}{6})+(\frac{45+4}{36})+...\)

\(=6+\frac{17}{6}+\frac{49}{36}+...\)

Since the given series is geometric series. So, use formula of geometric sum of infinite series

\(S_n=\frac{a(1-r^n)}{1-r}\)

where a is the first term of series and r is geometric ratio between every two terms.

As, the given series can be written as

\(I=\sum_{n=0}^\infty(\frac{5}{2^n})+\sum_{n=0}^\infty(\frac{1}{3^n})\)

If the series

\(\sum a_n=\sum_{n=0}^\infty(\frac{5}{2^n})\quad\text{and}\quad\sum b_n=\sum_{n=0}^\infty\frac{1}{3^n}\)

are convergent series then it can be written in the form of

\(\sum a_n+b_n=\sum a_n+\sum b_n\)

Check the convergence of the series by ratio test

\(\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow\infty}\frac{\frac{5}{2^{n+1}}}{\frac{5}{2^n}}\)

\(=\lim_{n\rightarrow\infty}\frac{2^n}{2^{n+1}}\)

\(=\lim_{n\rightarrow\infty}\frac{2^n}{2^n(2)}\)

\(=\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\frac{1}{2}<1\)</span>

As, the limit for above series is less than 1 . Therefore, the above series is convergent.

Now, check for another series

\(=\lim_{n\rightarrow\infty}\frac{b_{n+1}}{b_n}=\frac{\frac{1}{3^{n+1}}}{\frac{1}{3^n}}\)

\(=\lim_{n\rightarrow\infty}\frac{3^n}{3^n(3)}\)

\(=\lim_{n\rightarrow\infty}\frac{b_{n+1}}{b_n}=\frac{1}{3}<1\)</span>

It implies that the limit of above series is less than 1 and hence is convergent.

Therefore, the given series is

\(I=5\sum_{n=0}^\infty(\frac{1}{2^n})+\sum_{n=0}^{\infty}(\frac{1}{3^n})\)

By using formula of sum of infinite series

\(5(2)+\frac{3}{2}\)

\(\frac{23}{2}\)

Hence,the sum of given infinite series is \(\frac{23}{2}\)