Hidden Stuff

$\begin{aligned} S_n & = 1+3+5+7+\cdots+(2n-1) \\ & = \sum_{k=1}^n (2k-1) \\ & = \frac {2n(n+1)}2 - n \\ & = n^2 \end{aligned}$

For $2n-1=2017$ , $\implies n = 1009$ , $\implies S_{1009} = 1009^2$ , $\implies n = \boxed{\pm 1009}$

If you add one to the last number and divide by two, you will get n since all the numbers will be equal to 1009*1009 so the answer is 1009. If you want proof, $\sqrt{(2017+1)*(2018/4)}=\pm 1009$ .

Let $S$ be the sum of all positive odd integers up to $m$ . For instance, all positive odd integers up to 9. $S=1+3+5+7+9=25$ Let $k$ be the number of odd integers, then $S=k^2$ (I will not feature the proof in this solution). For example, if I want to find the sum of all odd positive integers up to $m$ , then $S=$ $(\dfrac{m+1}{2})^2$ . Here, $m=2017$ , hence $S=$ $(\dfrac{2017+1}{2})^2$ $=1018081$ $=n^2$

$n=\sqrt{1018081}=1009$

Let $a_{k}$ be an arithmetic sequence with common difference $d=2$ and first term $a_{1}=1$ . Then:
$a_{k}=a_{1}+d(k-1)\Rightarrow$
$a_{k}=2k-1$
Let i be the natural number for which:
$a_{i}=2017\Rightarrow$
$2i-1=2017\Rightarrow$
$i=1009$ .
The wanted sum is the sum of the first $i$ terms of $a_{k}$ which is:
$S_{i}=\frac{i}{2}(a_{1}+a_{i})\Rightarrow$
$S_{1009}=\frac{1009}{2}2018\Rightarrow$
$n^{2}=1009^{2}\Rightarrow$
$\boxed{n=\pm 1009}$