Given that,
\(n^{th}\ \text{term of the sequence is}\ 3^{n\ +\ 1}.\)
Take \(n = 0\)
\(a_{n} = 3^{n\ +\ 1}\)

\(a_{0} = 3^{0\ +\ 1} = 3\) Therefore, the first term is 3. Take \(n = 1,\)

\(a_{1} = 3^{1\ +\ 1}\)

\(a_{1} = 3^{1\ +\ 1} = 3^{2} = 9\) The second term is 9 Take \(n = 2,\)

\(a2 = 3^{2\ +\ 1}\)

\(= 3^{3}\)

\(a_{2} = 27\) The third term is 27 For, \(n = 3,\)

\(a3 = 3^{3\ +\ 1}\)

\(= 3^{4}\)

\(a^{3} = 81\) Fourth term is 81 For, \(n = 4,\)

\(a4 = 3^{4\ +\ 1}\)

\(= 3^{5}\)

\(a^{4} = 243\) Fifth term is 243. Therefore, the first 5 terms of the sequence are \(3,\ 9,\ 27,\ 81,\ 243.\) Step 3 The given sequence is \(3,\ 9,\ 27,\ 81,\ 243.\) The series associated with the given sequence is \(\sum_{n=0}^{5}\ 3^{n\ +\ 1}=3\ +\ 9\ +\ 27\ +\ 81\ +\ 23\)

\(\sum_{n=0}^{5}\ 3^{n\ +\ 1}=363\)

\(a_{0} = 3^{0\ +\ 1} = 3\) Therefore, the first term is 3. Take \(n = 1,\)

\(a_{1} = 3^{1\ +\ 1}\)

\(a_{1} = 3^{1\ +\ 1} = 3^{2} = 9\) The second term is 9 Take \(n = 2,\)

\(a2 = 3^{2\ +\ 1}\)

\(= 3^{3}\)

\(a_{2} = 27\) The third term is 27 For, \(n = 3,\)

\(a3 = 3^{3\ +\ 1}\)

\(= 3^{4}\)

\(a^{3} = 81\) Fourth term is 81 For, \(n = 4,\)

\(a4 = 3^{4\ +\ 1}\)

\(= 3^{5}\)

\(a^{4} = 243\) Fifth term is 243. Therefore, the first 5 terms of the sequence are \(3,\ 9,\ 27,\ 81,\ 243.\) Step 3 The given sequence is \(3,\ 9,\ 27,\ 81,\ 243.\) The series associated with the given sequence is \(\sum_{n=0}^{5}\ 3^{n\ +\ 1}=3\ +\ 9\ +\ 27\ +\ 81\ +\ 23\)

\(\sum_{n=0}^{5}\ 3^{n\ +\ 1}=363\)