Sequence Induction

Question

$$a_n=2a_{n-1}+3a _{n-2}$$ for $n\ge 3$

Given $a_1=a_2=1$, prove that$$a_n=\frac{1}{2}(3^{n-1}-(-1)^n)$$

Base Case: $a_1=a_2=1$

Induction Step($k\ge 3$): $$\implies a_n=2\cdot\frac{1}{2}(3^{k-1}-(-1)^k)+3\cdot\frac{1}{2}(3^{k-2}-(-1)^{k-1})$$ $$\implies a_n=3^{k-1}-(-1)^k+\frac{3}{2}\cdot3^{k-2}-\frac{3}{2}\cdot(-1)^{k-1}$$ $$\implies a_n=3^{k-1}+(-1)^{k+1}+\frac{1}{2}\cdot3^{k-1}+\frac{3}{2}(-1)^{k}$$ $$\implies a_n=\frac{3}{2}\cdot3^{k-1}+(-1)^{k+1}+\frac{3}{2}(-1)^{k}$$ $$\implies a_n=\frac{1}{2}\cdot3^{k}+(-1)^{k+1}+\frac{3}{2}(-1)^{k}$$

Im having trouble with the second term, i can't factor it out

How is this different from your other question yesterday? Just because the coefficients are not exactly the same doesn't make it a different question. — dxiv, Jul 16 '17 at 21:52
@dxiv i didnt realize they are the same and plus im not asking for all the help, i just needed help with the last term — TheGamer, Jul 16 '17 at 21:57
your first line must be $$2\cdot \frac{1}{2}(3^{n-2}-(-1)^{n-1})+3\cdot \frac{1}{2}\cdot(3^{k-3}-(-1)^{k-2})$$ — Dr. Sonnhard Graubner, Jul 16 '17 at 21:58
@Dr.SonnhardGraubner but is that goanna give me the final answer that i want — TheGamer, Jul 16 '17 at 22:02

score 2 · Accepted Answer · answered Jul 16 '17 at 21:35

2

Write $3/2(-1)^k$ as $-3/2(-1)^{k+1}$ and then the second and third term cancel down to the required result.

answered Jul 16 '17 at 21:35

PJF49

506

how did u get that term – TheGamer Jul 16 '17 at 21:36
@TheGamer It's the last term in your last line. – PJF49 Jul 16 '17 at 21:38
ik that but how did u come up with the $-\frac{3}{2}\cdot(-1)^{k+1}$ – TheGamer Jul 16 '17 at 21:39
@TheGamer I've multiplied the 3/2 by -1 and $(-1)^k$ by -1 and so have multiplied by -1 twice (so multiplied by 1). – PJF49 Jul 16 '17 at 21:49

Dr. Sonnhard Graubner · Answer 2 · 2017-07-16T21:38:37.600

1

one can do it like follows: $$\frac{1}{2}3^k-(-1)^k+\frac{3}{2}(-1)^k=\frac{1}{2}3^k+(-1)^k\left(\frac{3}{2}-1\right)=\frac{1}{2}\left(3^k+(-1)^k\right)$$ note that if $$a_n=\frac{1}{2}(3^{n-1}-(-1)^n)$$ then $$a_{n-1}=\frac{1}{2}(3^{n-2}-(-1)^{n-1})$$ and $$a_{n-2}=\frac{1}{2}(3^{n-3}-(-1)^{n-2})$$

edited Jul 16 '17 at 21:38

answered Jul 16 '17 at 21:30

Dr. Sonnhard Graubner

95,283

shouldnt it be $(3^k - (-1)^k)$, the answer in my textbook is $(3^k - (-1)^{k+1})$, – TheGamer Jul 16 '17 at 21:30
i have only used your last line – Dr. Sonnhard Graubner Jul 16 '17 at 21:53

Sequence Induction

2 Answers2