Given the series:

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

Rewriting the given series as sum of two series:

\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty\left(\frac{3^n}{4^n}+\frac{2^{n+1}}{4^n}\right)\)

\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty\frac{3^n}{4^n}+\sum_{n=0}^\infty\frac{2^{n+1}}{4^n}\)

\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty(\frac{3}{4})^n+\sum_{n=0}^\infty\frac{2\cdot2^n}{4^n}\)

\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty(\frac{3}{4})^n+\sum_{n=0}^\infty2(\frac{2}{4})^n\)

\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty(\frac{3}{4})^n+\sum_{n=0}^\infty2(\frac{1}{2})^n\)

Hence we get sum of two geometric series:

\(r_1=\frac34\)

\(r_2=\frac12\)

Since both \(|r_1|<1,|r_2|<1\)</span>

Both the individual geometric series converge.

The given series is sum of two converging, hence the given series \(\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}\) converges.

The sum of an infinite geometric series is given as:

\(S=\frac{\text{1st Term}}{1-r}\)

The sum of the series will be:

\(\sum_{n=0}^\infty(\frac34)^n=\frac{1}{1-\frac34}\)

\(\Rightarrow\sum_{n=0}^\infty(\frac34)^n=\frac{1}{\frac14}=4\)

\(\sum_{n=0}^\infty2(\frac12)^n=\frac{2}{\frac12}=4\)

The sum of the given series will be:

\(\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=4+4\)

\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=8\)

Final Answer:

The given series is converging.

The sum of series is 8

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

Rewriting the given series as sum of two series:

\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty\left(\frac{3^n}{4^n}+\frac{2^{n+1}}{4^n}\right)\)

\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty\frac{3^n}{4^n}+\sum_{n=0}^\infty\frac{2^{n+1}}{4^n}\)

\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty(\frac{3}{4})^n+\sum_{n=0}^\infty\frac{2\cdot2^n}{4^n}\)

\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty(\frac{3}{4})^n+\sum_{n=0}^\infty2(\frac{2}{4})^n\)

\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty(\frac{3}{4})^n+\sum_{n=0}^\infty2(\frac{1}{2})^n\)

Hence we get sum of two geometric series:

\(r_1=\frac34\)

\(r_2=\frac12\)

Since both \(|r_1|<1,|r_2|<1\)</span>

Both the individual geometric series converge.

The given series is sum of two converging, hence the given series \(\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}\) converges.

The sum of an infinite geometric series is given as:

\(S=\frac{\text{1st Term}}{1-r}\)

The sum of the series will be:

\(\sum_{n=0}^\infty(\frac34)^n=\frac{1}{1-\frac34}\)

\(\Rightarrow\sum_{n=0}^\infty(\frac34)^n=\frac{1}{\frac14}=4\)

\(\sum_{n=0}^\infty2(\frac12)^n=\frac{2}{\frac12}=4\)

The sum of the given series will be:

\(\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=4+4\)

\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=8\)

Final Answer:

The given series is converging.

The sum of series is 8