Question

1. find the first five terms of the sequence given by (a_1 = 2, a_n = 3a_{n - 1}-1). (3 points)
2. the first term of an arithmetic sequence is 2 and the fourth term is 11. find the sum of the first 50 terms. (3 points)

Explanation:

Step1: Find the first five - terms of the sequence \(a_n = 3a_{n - 1}-1\) with \(a_1 = 2\)

For \(n = 2\):
\[a_2=3a_1 - 1=3\times2 - 1=6 - 1 = 5\]

Step2: For \(n = 3\)

\[a_3=3a_2 - 1=3\times5 - 1=15 - 1 = 14\]

Step3: For \(n = 4\)

\[a_4=3a_3 - 1=3\times14 - 1=42 - 1 = 41\]

Step4: For \(n = 5\)

\[a_5=3a_4 - 1=3\times41 - 1=123 - 1 = 122\]
The first five - terms are \(2,5,14,41,122\).

Step5: For the arithmetic sequence problem

The formula for the \(n\)th term of an arithmetic sequence is \(a_n=a_1+(n - 1)d\). Given \(a_1 = 2\) and \(a_4=11\).
Substitute into the formula: \(a_4=a_1+(4 - 1)d\), so \(11 = 2+3d\).
Solve for \(d\):
\[3d=11 - 2=9\]
\[d = 3\]

Step6: The formula for the sum of the first \(n\) terms of an arithmetic sequence is \(S_n=\frac{n}{2}(2a_1+(n - 1)d)\).

Substitute \(n = 50\), \(a_1 = 2\), and \(d = 3\) into the formula:
\[S_{50}=\frac{50}{2}(2\times2+(50 - 1)\times3)\]
\[S_{50}=25(4 + 49\times3)\]
\[S_{50}=25(4+147)\]
\[S_{50}=25\times151\]
\[S_{50}=3775\]

Answer:

The first five - terms of the sequence \(a_n = 3a_{n - 1}-1\) with \(a_1 = 2\) are \(2,5,14,41,122\). The sum of the first 50 terms of the arithmetic sequence is \(3775\).