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

Feb 9, 2017

$89$

Explanation:

Label the terms of the sequence ${a}_{1} , {a}_{2} , \ldots , {a}_{2016} , \ldots$.

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 $2 k - 1$ is ${a}_{{S}_{k}}$ where

${S}_{k} = {\sum}_{i = 1}^{k} \left(2 i - 1\right) = 1 + 3 + 5 + 7 + \ldots + \left(2 k - 1\right)$

Then, to know the value of the terms in the subsequence containing ${a}_{2016}$, we need only find $k$ such that ${S}_{k - 1} < 2016 \le {S}_{k}$. Then we will have ${a}_{2016} = 2 k - 1$. To do so, we will put ${S}_{k}$ into a closed form.

${S}_{k} = {\sum}_{i = 1}^{k} \left(2 i - 1\right)$

$= - k + 2 {\sum}_{i = 1}^{k} i$

$= - k + 2 \left(\frac{k \left(k + 1\right)}{2}\right)$

$= - k + k \left(k + 1\right)$

$= {k}^{2}$

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

Examining squares, we find that

${S}_{44} = {44}^{2} < 2016 \le {45}^{2} = {S}_{45}$

Thus ${a}_{2016}$ occurs after the final term of the subsequence consisting of $2 \left(44\right) - 1$ but before the last term of the subsequence consisting of $2 \left(45\right) - 1$, giving us ${a}_{2016} = 89$.