Infinite sequence

In summation form, $S$ is as follows:

$\begin{aligned} S & = \sum_{n=0}^\infty \frac {a_n}{10^{2n}} = \sum_{n=0}^\infty \frac {a_n}{100^n} \\ & = \frac {a_0}{100^0} + \frac {a_1}{100^1} + \sum_{n=2}^\infty \frac {a_n}{100^n} \\ & = 1 + \frac 1{100} + \sum_{n=2}^\infty \frac {20a_{n-1}+12a_{n-2}}{100^n} \\ & = \frac {101}{100} + 20 \sum_{\red{n=1}}^\infty \frac {a_n}{100^{n+1}} + 12 \sum_{n=0}^\infty \frac {a_n}{100^{n+2}} \\ & = \frac {101}{100} + \frac {20}{100} \left(\sum_{\red{n=0}}^\infty \frac {a_n}{100^n} - \red{\frac {a_0}{100^0}} \right) + \frac {12}{10000} \sum_{n=0}^\infty \frac {a_n}{100^n} \\ & = \frac {101}{100} + \frac 15 S - \frac 15 + \frac 3{2500} S \\ \left(1-\frac 15 - \frac 3{2500} \right) S & = \frac {101}{100} - \frac 15 \\ \frac {1997}{2500} S & = \frac {81}{100} \\ S & = \frac {81 \times 25}{1997} \\ \implies \sqrt S & = \frac {45}{\sqrt{1997}} \end{aligned}$

Therefore $a+b = 45+1997 = \boxed{2042}$ .

Let

$f\left(x\right)=\sum_{n=0}^{\infty}a_{n}x^{n}$

Multiplying the reccurence relation by $x^{n}$ and summing the relation from $n=2$ to $n=\infty$ we get

$f\left(x\right)-a_{1}x-a_{0}=20x\left(f\left(x\right)-a_{0}\right)+12x^{2}f\left(x\right)$

Putting $\displaystyle a_{0}=a_{1}=1$ and then after manipulating a bit, we finally arrive at

$f\left(x\right)=\frac{19x-1}{12x^{2}+20x-1}$

Now, simply observe that $\displaystyle S=f\left(10^{-2}\right)=\frac{2025}{1997}$ and thus we arrive at our answer .

$\begin{aligned}\displaystyle S-\frac{20S}{10^2}-\frac{12S}{10^4} &= (\frac{a_0}{10^0}+\frac{a_1}{10^2}+\frac{a_2}{10^4}+\frac{a_3}{10^6}+\dots)-(\frac{20a_0}{10^2}+\frac{20a_1}{10^4}+\frac{20a_2}{10^6}+\frac{20a_3}{10^8}+\dots)-(\dfrac{12a_0}{10^4}+\dfrac{12a_1}{10^6}+\dfrac{12a_2}{10^8}+\dfrac{12a_3}{10^{10}}+\dots) \\ \displaystyle &=\frac{a_0}{10^0}+\frac{a_1}{10^2}-\frac{20a_0}{10^2}+\frac{a_2-20a_1-12a_0}{10^4}+\frac{a_3-20a_2-12a_1}{10^6}+\frac{a_4-20a_3-12a_2}{10^8}+\dots \end{aligned}$

Since $a_n=20a_{n-1}+12a_{n-2}$ for all positive integers $n\geq2$ , so

$\displaystyle S-\frac{20S}{10^2}-\frac{12S}{10^4}=\frac{a_0}{10^0}+\frac{a_1}{10^2}-\frac{20a_0}{10^2}$

and as $a_0=a_1=1$ , we have

$\displaystyle \frac{7988S}{10000}=\frac{81}{100}$

$\displaystyle \implies \sqrt{S}=\frac{45}{\sqrt{1997}}$

$\boxed{\therefore a+b=45+1997=2042}$