Question

which recursive sequence would produce the sequence 3, 11, 43, ...?
answer attempt 1 out of 2
○ $a_1 = 3$ and $a_n = 4a_{n - 1} - 1$
○ $a_1 = 3$ and $a_n = -a_{n - 1} + 4$
○ $a_1 = 3$ and $a_n = 3a_{n - 1} + 2$
○ $a_1 = 3$ and $a_n = 2a_{n - 1} + 3$
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Explanation:

Step1: Test the first option

Given \(a_1 = 3\) and \(a_n=4a_{n - 1}-1\). For \(n = 2\), \(a_2=4a_1 - 1=4\times3-1 = 11\). For \(n = 3\), \(a_3=4a_2-1=4\times11 - 1=43\). This matches the sequence.

Step2: Test the second option

Given \(a_1 = 3\) and \(a_n=-a_{n - 1}+4\). For \(n = 2\), \(a_2=-3 + 4=1
eq11\). So this option is wrong.

Step3: Test the third option

Given \(a_1 = 3\) and \(a_n=3a_{n - 1}+2\). For \(n = 2\), \(a_2=3\times3+2 = 11\). For \(n = 3\), \(a_3=3\times11+2=35
eq43\). So this option is wrong.

Step4: Test the fourth option

Given \(a_1 = 3\) and \(a_n=2a_{n - 1}+3\). For \(n = 2\), \(a_2=2\times3+3 = 9
eq11\). So this option is wrong.

Answer:

A. \(a_1 = 3\) and \(a_n = 4a_{n-1}-1\)