jmc

Question

The sequence a_0, a_1, a_2, , satisfies the recurrence equation a_n = 2 a_n-1 - 2 a_n - 2 + a_n - 3 for every integer n 3. If a_20 = 1, a_25 = 10, and a_30 = 100, what is the value of a_1331?

Accepted Answer

We can calculate the first few terms of the sequence and look for a pattern. For

n = 3

,

a_{3} = 2 a_{2} - 2 a_{1} + a_{0} .

For

n = 4

we get

a_{4} = 2 a_{3} - 2 a_{2} + a_{1} = 2 (2 a_{2} - 2 a_{1} + a_{0}) - 2 a_{2} + a_{1} = 2 a_{2} - 3 a_{1} + 2 a_{0} .

With

n = 5

we have

a_{5} = 2 a_{4} - 2 a_{3} + a_{2} = 2 (2 a_{2} - 3 a_{1} + 2 a_{0}) - 2 (2 a_{2} - 2 a_{1} + a_{0}) + a_{2} = a_{2} - 2 a_{1} + 2 a_{0} .

With

n = 6

we have

a_{6} = 2 a_{5} - 2 a_{4} + a_{3} = 2 (a_{2} - 2 a_{1} + 2 a_{0}) - 2 (2 a_{2} - 3 a_{1} + 2 a_{0}) + 2 (2 a_{2} - 2 a_{1} + a_{0}) = a_{0} .

Brilliant! We found that

a_{6} = a_{0}

and we can similarly check that

a_{7} = a_{1}

and

a_{8} = a_{2}

and so on because of the recursive rules of the sequence. This means that the sequence is periodic with a period of 6.

This means that

a_{0} = a_{30} = 100

. Similarly,

a_{1} = a_{25} = 10

and

a_{2} = a_{20} = 1

. Then,

a_{1331} = a_{5} = a_{2} - 2 a_{1} + 2 a_{0} = 1 - 2 (10) + 2 (100) = 181 .