Discrete Math: Induction and Recurrence. Show that the terms of the sequence that satisfy a_n=3a_{n-1}-2 and a_1=4 are given by the formula a_n=3^n+1 for all n >= 1.

Kendra Hudson

Kendra Hudson

Answered question

2022-09-04

Discrete Math: Induction and Recurrence
Show that the terms of the sequence that satisfy a n = 3 a n 1 2 and a 1 = 4 are given by the formula a n = 3 n + 1 for all n 1.
I know this problem has to do with induction. In this case I think n = n + 1 so I would have 3 n + 1 but i'm not sure what to do with the rest of the problem. Can someone help please?

Answer & Explanation

Holly Schmidt

Holly Schmidt

Beginner2022-09-05Added 10 answers

Step 1
I guess the problem is
a n = 3 a n 1 2 a n 1 = 3 ( a n 1 1 )
Writing b n = a n 1 ,, we get b n = 3 b n 1 which implies b n = 3 m b n m
Step 2
Now, b 1 = a 1 1 = 3 , , and setting m = n 1 b n = 3 n 1 b 1 =
incibracy5x

incibracy5x

Beginner2022-09-06Added 21 answers

Step 1
a 2 = 3 a 1 2 = 3 4 2 = 10. I've solved a very similar question that was posted here yesterday that required the generating function approach. But the characteristic equation trick seems to work well for this question also. So x 2 3 x = 0 x ( x 3 ) = 0 x = 0 , 3. So the general solution would be: a n = p 0 n + q 3 n + r = q 3 n + r. Using the initial conditions we obtain: 4 = a 1 = 3 q + r, and
10 = a 2 = 9 q + r 4 3 q = 10 9 q 6 q = 6 q = 1
and 4 = 3 1 + r r = 1.
Step 2
Thus:
a n = 1 3 n + 1 = 3 n + 1 , n 1

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