Prove $a_n =3\cdot2^{n-1} +2(-1)^n$

Question

$a_1=1, a_2=8$ and $a_n=a_{n-1} +2a_{n-2}$ for $n\ge3$

Prove that

$$a_n =3\cdot2^{n-1} +2(-1)^n$$ for $n\in N$

Base Case: n=1

$a_1= 3\cdot2^0+2(-1)^1, a_1 =1$, the base case holds true

I.H: Suppose its true for $$a_k =3\cdot2^{k-1} +2(-1)^k$$

How do i proceed for induction step, am i suppose to use $a_n=a_{n-1} +2a_{n-2}$ in the induction step?

Guide:
check that it is true for $a_2$ as well.

now you can suppose it is true for $n=k-1$ and $n=k-2$ and prove that it is true for $n=k$. — Siong Thye Goh, Jul 15 '17 at 23:00
@SiongThyeGoh yes it works for $a_2$ and how did u get n=k-1 and n=k-2 — TheGamer, Jul 15 '17 at 23:01
That is the induction hypothesis. Suppose $a_{k-1}=3.2^{(k-1)-1}+2(-1)^{k-1}$ and $a_{k-2}=3.2^{(k-2)-1}+2(-1)^{k-2}$, try to prove that $a_{k}=3.2^{(k)-1}+2(-1)^{k}$ — Siong Thye Goh, Jul 15 '17 at 23:04

score 1 · Accepted Answer · edited Jul 15 '17 at 23:52

1

NOTE: For this problem, you need two base cases, since the recursive sequence is defined in terms of the previous two terms. I'll leave that part up to you.

INDUCTION STEP: Assume that both $$a_n=3(2^{n-1})+2(-1)^n$$ and $$a_{n+1}=3(2^{n})+2(-1)^{n+1}$$ for some $n\ge 3$. Then, by hypothesis $$a_{n+2}=a_{n+1}+2a_n$$ so substituting the values of $\,a_{n+1}\,$ and $\,a_{n}$ from our induction hypothesis yields $$\begin{align} a_{n+2}&=3(2^{n})+2(-1)^{n+1}+2(3(2^{n-1})+2(-1)^n)\\ &=3(2^{n})+2(-1)^{n+1}+3(2^{n})+4(-1)^n\\ &=6(2^{n})+2(-1)^{n}\\ &=3(2^{n+1})+2(-1)^{n+2} \end{align}$$ which completes the proof of the inductive step.

edited Jul 15 '17 at 23:52

Bill Dubuque

272,048

answered Jul 15 '17 at 23:04

Franklin Pezzuti Dyer

39,754
9
73
166

im confused why didnt u prove n+1 and how did u come up with that equation $a_{n+2}=a_{n+1}+2a_n$. Also why is Siong saying to prove for n=k-1 and n=k-2 – TheGamer Jul 15 '17 at 23:15
@TheGamer I did prove that, but I just "shifted it over". Instead of proving it for all $n+1$, I proved it for all $n+2$, which is basically the same. Also, I got $a_{n+2}=a_{n+1}+2a_n$ because it is one of the givens of your problem. Finally, you need to have two base cases since the sequence is defined recursively in terms of the previous two terms. – Franklin Pezzuti Dyer Jul 15 '17 at 23:18
y ru doing n+1, isnt it supposed to be n-1 and $a_n=a_{n-1} +2a_{n-2}$ this doesnt give u $a_{n+2}=a_{n-1} +2a_n$ – TheGamer Jul 15 '17 at 23:20
@TheGamer What part of my answer are you referring to? – Franklin Pezzuti Dyer Jul 15 '17 at 23:21
@TheGamer No, it should not. – Franklin Pezzuti Dyer Jul 15 '17 at 23:25
@TheGamer Do you know Complete / Strong Induction? That makes proofs like this clearer, e.g. see here. – Bill Dubuque Jul 15 '17 at 23:31
1

I made minor changes to the exponents in the 2nd terms of the last two lines, but feel free to correct my changes if they are in error. – user84413 Jul 15 '17 at 23:40
i get everything except how u came up with that equation $a_{n+2}=a_{n+1}+2a_n$ – TheGamer Jul 15 '17 at 23:42
@BillDubuque ik for induction, u assume its true for P(K) and u goanna show its true for P(k+1) – TheGamer Jul 15 '17 at 23:44
@TheGamer It is a given that $a_n=a_{n-1}+2a_{n-2}$, correct? – Franklin Pezzuti Dyer Jul 15 '17 at 23:45
yes thats right – TheGamer Jul 15 '17 at 23:45
@TheGamer From that equation, by substituting $n\to n+2$, it follows that $$a_{n+2}=a_{n+1}+2a_n$$ – Franklin Pezzuti Dyer Jul 15 '17 at 23:46
ohhh, wow also rather than doing n+2, can i just do it for n+1, so its just $a_{n+1}= a_n +2a_{n-1}$ – TheGamer Jul 15 '17 at 23:50
@TheGamer Right. Did you find my answer helpful? – Franklin Pezzuti Dyer Jul 15 '17 at 23:53
@Nilknarf can i do that or no – TheGamer Jul 15 '17 at 23:55
@TheGamer Yes, that's right. In fact, you could even go so far as to say $$a_{n+101}=a_{n+100}+2a_{n+99}$$! – Franklin Pezzuti Dyer Jul 15 '17 at 23:55
@TheGamer The inductive hypothesis $P(n)$ is that the equation is true at a pair of neighboring integers $n, n+1$. So $P(n+1)$ means it is true at $n+1, n+2,,$ Since it is true at $n+1$ by (induction) hypothesis, we only need to prove it is true at $n+2$ to complete the inductive step - exactly what is done in this answer. If you study complete / strong / course-of-values induction then you will gain a better understanding of this type of induction (above is a special case). – Bill Dubuque Jul 15 '17 at 23:57
@BillDubuque Im stuck on this part $6\cdot2^{n-1} +2(-1)^n +4(-1)^{n-1}$ – TheGamer Jul 16 '17 at 02:22
@TheGamer The first paragraph in my answer shows how to rewrite the last two terms to $2(-1)^{n+1}$ and the first term $,6\cdot 2^{n-1} = 3\cdot 2^n\ $ – Bill Dubuque Jul 16 '17 at 02:24
but is that same as $3\cdot2^{n-1}$ – TheGamer Jul 16 '17 at 02:27

score 0 · Answer 2 · answered Jul 16 '17 at 01:43

Let's consider two real $r_1,r_2$ of sum $s$ and product $p$ then if $(a_n)_n$ verifies

$a_{n}-s\,a_{n-1}+p\,a_{n-2}=0$

Then $\exists\ \alpha,\beta$ such that $a_n=\alpha(r_1)^n+\beta(r_2)^n$.

It can be proved by induction, if we assume the formula verified for initial terms then

$\require{cancel} a_{n+1}=sa_{n}-pa_{n-1}=(r_1+r_2)\left(\alpha r_1^n+\beta r_2^n\right)-r_1r_2\left(\alpha r_1^{n-1}+\beta r_2^{n-1}\right)=\alpha r_1^{n+1}+\cancel{\alpha r_2r_1^n}+\cancel{\beta r_1r_2^n}+\beta r_2^{n+1}-\cancel{\alpha r_2r_1^n}-\cancel{\beta r_1r_2^n}=\alpha r_1^{n+1}+\beta r_2^{n+1}$

So whe you have an equation of the type $a_n=a_{n-1}+2a_{n-2}$

Then you start by solving $r^2=r+2\iff r^2-r-2=0\iff(r+1)(r-2)=0$

Whose roots are $r_1=-1$ and $r_2=2$.

So there exist $\alpha,\beta$ such that $a_n=\alpha(-1)^n+\beta(2^n)$

Initial terms give $\begin{cases} a_1=1=-\alpha+2\beta\\ a_2=8=\alpha+4\beta\end{cases}\iff\begin{cases}\alpha=2\\\beta=\frac 32\end{cases}$

And finally $a_n=2(-1)^n+\frac 32\cdot2^n=2(-1)^n+3\cdot2^{n-1}$

You can find more on this general method here :

https://en.wikipedia.org/wiki/Linear_difference_equation

One can also view this in terms of linear shift operators, i.e. write the recurrence as $,(S-2)(S+1) a_n = 0,$ where $S a_n = a_{n+1}$, e.g. see this answer. — Bill Dubuque, Jul 16 '17 at 01:50

score -1 · Answer 3 · answered Jul 15 '17 at 23:01

-1

hint

Let for $n\ge 3$, $$b_n=a_n+a_{n-1} $$

then $$b_n=2b_{n-1} $$ Geometric.

answered Jul 15 '17 at 23:01

hamam_Abdallah

62,951

You sure meant $a_{n+1} - a_{n}$ – Alex Jul 16 '17 at 01:44
@Alex No, it is correct as written: $,a_n\color{red}{+a_{n-1}}=a_{n-1} +2a_{n-2}\color{red}{+a_{n-1}}=2(a_{n-1}+a_{n-2})$ so $b_n = 2b_{n-1},$. – dxiv Jul 16 '17 at 04:51
P.S. I don't understand the downvotes. – dxiv Jul 16 '17 at 04:52

Prove $a_n =3\cdot2^{n-1} +2(-1)^n$

3 Answers3

Linked