We are given a sequence $\{a_n\}$ with the initial term $a_1 = 2$ and the recursive relation $a_{n+1} = 3a_n + 4$. We want to find the general term $a_n$ of the sequence.

Discrete MathematicsSequencesRecurrence RelationsGeometric Sequence
2025/3/6

1. Problem Description

We are given a sequence {an}\{a_n\} with the initial term a1=2a_1 = 2 and the recursive relation an+1=3an+4a_{n+1} = 3a_n + 4. We want to find the general term ana_n of the sequence.

2. Solution Steps

Let's try to transform the recursive relation into a more manageable form.
Assume that an+1+x=3(an+x)a_{n+1} + x = 3(a_n + x) for some constant xx. Then an+1=3an+3xxa_{n+1} = 3a_n + 3x - x, so an+1=3an+2xa_{n+1} = 3a_n + 2x.
Comparing this with an+1=3an+4a_{n+1} = 3a_n + 4, we have 2x=42x = 4, so x=2x = 2.
Thus, an+1+2=3(an+2)a_{n+1} + 2 = 3(a_n + 2).
Let bn=an+2b_n = a_n + 2. Then bn+1=3bnb_{n+1} = 3b_n, which means that {bn}\{b_n\} is a geometric sequence with a common ratio of 33.
We have b1=a1+2=2+2=4b_1 = a_1 + 2 = 2 + 2 = 4.
The general term for bnb_n is given by bn=b13n1b_n = b_1 \cdot 3^{n-1}.
bn=43n1b_n = 4 \cdot 3^{n-1}
Since bn=an+2b_n = a_n + 2, we have an=bn2a_n = b_n - 2.
an=43n12a_n = 4 \cdot 3^{n-1} - 2
We can check the first few terms.
a1=43112=4302=412=2a_1 = 4 \cdot 3^{1-1} - 2 = 4 \cdot 3^0 - 2 = 4 \cdot 1 - 2 = 2
a2=3a1+4=3(2)+4=10a_2 = 3a_1 + 4 = 3(2) + 4 = 10. Also, a2=43212=432=122=10a_2 = 4 \cdot 3^{2-1} - 2 = 4 \cdot 3 - 2 = 12 - 2 = 10.
a3=3a2+4=3(10)+4=34a_3 = 3a_2 + 4 = 3(10) + 4 = 34. Also, a3=43312=492=362=34a_3 = 4 \cdot 3^{3-1} - 2 = 4 \cdot 9 - 2 = 36 - 2 = 34.

3. Final Answer

an=43n12a_n = 4 \cdot 3^{n-1} - 2

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