Reach for the Summit - M-S3-A1

Note that $a_1=S_1$ , and from the defining statement:

$\frac{a_1+2}{2}=\sqrt{2S_1}=\sqrt{2a_1},$

So then

$\begin{aligned}a_1+2=\sqrt{8a_1}\implies (a_1+2)^2&=8a_1\\\iff a_1^2-4a_1+4&=0\\\iff (a_1-2)^2&=0\\\iff a_1&=2\end{aligned}$

Continue like this to find a few terms, for example:

$\begin{aligned}a_2+2=\sqrt{8(2+a_2)}\implies (a_2+2)^2&=8(a_2+2)\\\iff a_2&=-2 \textnormal{ or }a_2=6\\\iff a_2&=6\because a_n>0\end{aligned}$

We obtain $\{a_n\}=\{2, 6, 10, 14, ...\}$ , so guess that $a_n=4n-2$ .

Proof that $a_n=4n-2$ :

$\begin{aligned}\frac{a_n+2}{2}&=\sqrt{2S_{n}}\\\iff a_n+2&=\sqrt{8S_{n}}\\\implies a_n^2+4a_n+4&=8S_{n}\\\iff S_n&=\frac{a_n^2+4a_n+4}{8}\ (*)\end{aligned}$

Fix $a_n=4n-2$ , so that

$\begin{aligned}S_n&=\sum_{r=1}^n 4n-2\\&=4\sum_{r=1}^n n-2\sum_{r=1}^n 1\\&=4\left (\frac{n(n+1)}{2}\right )-2n\\&=2(n^2+n)-2n\\&=2n^2\\\\\frac{a_n^2+4a_n+4}{8}&=\frac{(4n-2)^2+4(4n-2)+4}{8}\\&=\frac{(16n^2-16n+4)+(16n-8)+4}{8}\\&=\frac{16n^2}{8}\\&=2n^2\end{aligned}$

So $a_n=4n-2$ satisfies $(*)\ \square$ .

$\therefore a_{2020}=4\times 2020-2=8080-2=\color{#20A900}{\boxed{8078}}$ .