# B What is the formula for this sequence for n-th number

#### HAF

Hello, i have a sequence {1,2,13,62,313...} and I have to find out the rule for n-th number. I've found out that every next number is five times bigger but then is added or subtracted 3. For example 1x5 -3 = 2 and 2x5 +3 = 13 and so on. Can you please give me some advice how to create the general formula of this sequence?

Thank you

#### DrClaude

Mentor
I'll give you a hint that alternating signs can be obtained by raising -1 to a power that depends on n.

#### mfb

Mentor
It might also be interesting to compare this sequence with one where you only multiply by 5 each time.

#### WWGD

Gold Member
Hello, i have a sequence {1,2,13,62,313...} and I have to find out the rule for n-th number. I've found out that every next number is five times bigger but then is added or subtracted 3. For example 1x5 -3 = 2 and 2x5 +3 = 13 and so on. Can you please give me some advice how to create the general formula of this sequence?

Thank you
It is also possible to just define it by cases, one for when sequence term is even and one where term is odd.

#### jbriggs444

Homework Helper
It is also possible to just define it by cases, one for when sequence term is even and one where term is odd.
If the solution is of the form $a(-1^n) + b(5^n)$ then one should be able to find a characteristic equation for it -- a quadratic with roots of -1 and 5. That characteristic equation would then suggest a recurrence relation. Which immediately yields a recursive rule for the n'th number in terms of the n-1'st and n-2'nd.

Yup. Works out quite easily. [It's been almost 40 years since I learned how to go from a recurrence relation to a formula. This is the first time I've gone the other way -- from a formula to a recurrence relation]

• WWGD

#### WWGD

Gold Member
If the solution is of the form $a(-1^n) + b(5^n)$ then one should be able to find a characteristic equation for it -- a quadratic with roots of -1 and 5. That, characteristic equation would then suggest a recurrence relation. Which immediately yields a recursive rule for the n'th number in terms of the n-1'st and n-2'nd.

Yup. Works out quite easily.
Yes, I mean, my description may not be the best by reasonable standards, but it does describe the sequence fully.

• jbriggs444