SSome Help's question at Yahoo Answers regarding a linear recurrence equation

[MarkFL]

Here is the question:

Discrete Math Help questions?

Let {asubn} be the sequence defined recursively by asub1=11, asub2=37, asubn= 7asub(n-1) - 10asub(n-2) for n greater than or equal to 3.

a) find asub1, asub2, asub3, asub4.
b)Prove that asubn = 5^n + 3(2^n) for all n greater than or equal to 1.

Here is a link to the question:

Discrete Math Help questions? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Re: SSome Help's question at Yahoo! Answers regarding a linear reccurence equation

Hello SSome Help,

We are given the recursion:

\(\displaystyle a_n=7a_{n-1}-10a_{n-2}\) where \(\displaystyle a_1=11,\,a_2=37\)

The associated characteristic equation is:

\(\displaystyle r^2-7r+10=(r-2)(r-5)=0\)

Hence, the closed-form expression will take the form:

\(\displaystyle a_n=k_12^n+k_25^n\) where \(\displaystyle n\in\mathbb{N}\)

where the parameters $k_i$ may be determined from the initial values:

\(\displaystyle a_1=2k_1+5k_2=11\)

\(\displaystyle a_2=4k_1+25k_2=37\)

Solving this system, we find:

\(\displaystyle k_1=3,\,k_2=1\)

and so we have:

\(\displaystyle a_n=3\cdot2^n+5^n\)

a) We are given $a_1=11$ and $a_2=37$.

Using the recursive definition we find:

\(\displaystyle a_3=7\cdot37-10\cdot11=149\)

\(\displaystyle a_4=7\cdot149-10\cdot37=673\)

Using the closed-form we derived, we find:

\(\displaystyle a_3=3\cdot2^3+5^3=149\)

\(\displaystyle a_4=3\cdot2^4+5^4=673\)

b) We have already derived the closed-forum for the recursion.

To SSome Help and any other guests viewing this topic, I invite and encourage you to post other discrete math questions in our http://www.mathhelpboards.com/f15/ forum.

Best Regards,

Mark.