Question

Consider the sequence 1, 3, 3, 3, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 7, ... What is its 2016th term?

Answer 1

`89`

Explanation:

Label the terms of the sequence `a_1, a_2, ..., a_2016, ...`.

Notice that the last term with the value `1` is `a_1`, the last term with the value `3` is `a_4`, the last term with the value `5` is `a_9`, and in general, the last term with the value `2k-1` is `a_(S_k)` where

`S_k = sum_(i=1)^k(2i-1) = 1+3+5+7+...+(2k-1)`

Then, to know the value of the terms in the subsequence containing `a_2016`, we need only find `k` such that `S_(k-1) < 2016 <= S_k`. Then we will have `a_2016 = 2k-1`. To do so, we will put `S_k` into a closed form.

`S_k = sum_(i=1)^k(2i-1)`

`=-k+2sum_(i=1)^ki`

`=-k+2((k(k+1))/2)`

`=-k+k(k+1)`

`=k^2`

(That the sum of the first `n` odd integers is `n^2` is a useful result to remember)

Examining squares, we find that

`S_44 = 44^2 < 2016 <= 45^2 = S_45`

Thus `a_2016` occurs after the final term of the subsequence consisting of `2(44)-1` but before the last term of the subsequence consisting of `2(45)-1`, giving us `a_2016 = 89`.