In the sequence

, each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is

. What is the

th term in this sequence?

I know that nobody cares, but here’s my solution anyways.

This problem defines a linear recurrence such that:

Replacing with a constant in order to find the characteristic polynomial, :

Thus the characteristic polynomial of this linear recurrence is:

Rearrange, and divide by .

This is our characteristic polynomial. Factoring it:

Difference of cubes.

Combine like terms and factor .

The roots of the characteristic polynomial are . The root has a multiplicity of two, so the general solution to the recurrence is given by:

We can plug in the initial values given in the problem, , , , and to find the constants.

Solving this system of three linear equations gives the following values for , , and :

Therefore, the formula for the th term of this sequence is given by:

And, finally, is:

Hey, Andy, you must have solved it intuitively (instead of actually solving the recurrence and finding the general formula), could ya tell me/us how? =D

### Related Posts:

Well… Not really…. haha…

I solved like this:

First 4 terms: 2001, 2002, 2003, 2000

Solve for the next 2 terms….

X5 = 2002+2003-2000 = 2005

X6 = 2003+2000-2005 = 1998

U can find a pattern in every other term:

X1 = 2001

X3 = 2003

X5 = 2005

and

X2 = 2002

X4 = 2000

X6 = 1998

so you can divide the main sequence into two separate ones:

Sn (for X2, X4, X6) = 2002, 2000, 1998, 1996 ….

Sm(for X1, X3, X5) = 2001, 2003, 2005, 2007 ….

we really don’t need Sm to find the 2004th term so we’ll focus on Sn

Assuming the original sequence terminates, then the number of Sn is half of the the number of terms in the original sequence.

So… the 2004th term of the original sequence is the 1002th term of Sn

Sn is a simple arithmetic sequence, decreasing by 2 per element/number

thus term n = (term 1) + (n-1)(d)

or = 2002 – 2(n-1)

or = 2004 -2n.

Solve for n=1002,

=2004-2(1002)

=0

There’s probably an even easier way tho… buts thats how i solved it…… haha

=)

… that makes me feel extremely dumb for actually solving the whole recurrence, complete with finding the general formula and all.

…

>____<