Question

A sequence `{a_n}` is defined recursively, with `a_1 = 1`, `a_2=2` and, for `n>2`, `a_n = (a_(n-1))/(a_(n-2))`. How do you find the term `a_(241)`?

Answer 1

`a_241 = 1`

Explanation:

First let's observe some values of the sequence to find a common trend.

`a_3 = (a_2)/(a_1) = 2/1 = 2`
`a_4 = (a_3)/(a_2) = 2/2 = 1`
`a_5 = (a_4)/(a_3) = 1/2`
`a_6 = (a_5)/(a_4) = (1/2)/1=1/2`
`a_7 = (a_6)/(a_5) = (1/2)/(1/2) = 1 = a_1`
`a_8 = (a_7)/(a_6) = 1/(1/2) = 2 = a_2`

As you probably notice now, the sequence repeats every `6` terms, meaning that `a_1 = a_7 = a_13 = ... a_(1+6t)`.

In fact, we can generalize. If `1<=k<7`, then

`a_k = a_(k+6) = a_(k+12) = ... = a_(k+6t)`.

What this means that for any `n`, we can find which value `a_n` corresponds to by taking the remainder when we divide `n` by `6`.

In the case of `n=241`, `241/6 = (240+1)/6 = 40 1/6`, which implies a remainder of `1` and thus `a_241` corresponds to `a_1`, which is `1`.