Inspired by Yan Shee

$\displaystyle\sum_{n=1}^{\infty}\dfrac{n^2+n}{2\cdot10^n}=\displaystyle\sum_{n=1}^{\infty}\left(\dfrac12\cdot\dfrac{n^2+n}{10^n}\right)=\dfrac12\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{n^2}{10^n}+\dfrac{n}{10^n}\right)=\dfrac12\left(\color{#20A900}{\displaystyle\sum_{n=1}^{\infty}\dfrac{n^2}{10^n}}+\color{teal}{\displaystyle\sum_{n=1}^{\infty}\dfrac{n}{10^n}}\right)$

Let $\color{#20A900}{A=\displaystyle\sum_{n=1}^{\infty}\dfrac{n^2}{10^n}=\dfrac1{10}+\dfrac4{10^2}+\dfrac9{10^3}+\dfrac{16}{10^4}+\cdots}\Rightarrow(1)$

Multiplying both sides with $\dfrac1{10}$

$\dfrac1{10}A=\dfrac1{10^2}+\dfrac4{10^3}+\dfrac9{10^4}+\dfrac{16}{10^5}+\cdots\Rightarrow(2)$

Subtracting equation $(1)$ with equation $(2)$

$\dfrac9{10}A=\dfrac1{10}+\dfrac3{10^2}+\dfrac5{10^3}+\dfrac7{10^4}+\cdots\Rightarrow(3)$

Multiplying both sides with $\dfrac1{10}$

$\dfrac9{100}A=\dfrac1{10^2}+\dfrac3{10^3}+\dfrac5{10^4}+\dfrac7{10^5}+\cdots\Rightarrow(4)$

Subtracting equation $(3)$ with equation $(4)$

$\begin{aligned}\dfrac{81}{100}A=&\dfrac1{10}+\dfrac2{10^2}+\dfrac2{10^3}+\dfrac2{10^4}+\cdots\\\dfrac{81}{100}A=&\dfrac1{10}+\dfrac{\frac2{10^2}}{1-\frac1{10}}\\\dfrac{81}{100}A=&\dfrac1{10}+\left(\dfrac1{50}\times\dfrac{10}9\right)\\\dfrac{81}{100}A=&\dfrac1{10}+\dfrac1{45}\\\color{#20A900}{A=}&\color{#20A900}{\dfrac{9+2}{90}\times\dfrac{100}{81}=\dfrac{110}{729}}\end{aligned}$

Let $\color{teal}{B=\displaystyle\sum_{n=1}^{\infty}\dfrac{n}{10^n}=\dfrac1{10}+\dfrac2{10^2}+\dfrac3{10^3}+\dfrac4{10^4}+\cdots}\Rightarrow(5)$

Multiplying both sides with $\dfrac1{10}$

$\dfrac1{10}B=\dfrac1{10^2}+\dfrac2{10^3}+\dfrac3{10^4}+\dfrac4{10^5}+\cdots\Rightarrow(6)$

Subtracting equation $(5)$ with equation $(6)$

$\begin{aligned}\dfrac9{10}B=&\dfrac1{10}+\dfrac1{10^2}+\dfrac1{10^3}+\dfrac1{10^4}+\cdots\\\dfrac9{10}B=&\dfrac{\frac1{10}}{1-\frac1{10}}\\\color{teal}{B=}&\color{teal}{\dfrac1{10}\times\dfrac{10}9\times\dfrac{10}9=\dfrac{10}{81}}\end{aligned}$

Thus, $\displaystyle\sum_{n=1}^{\infty}\dfrac{n^2+n}{2\cdot10^n}=\dfrac12\left(\color{#20A900}{\dfrac{110}{729}}+\color{teal}{\dfrac{10}{81}}\right)=\dfrac12\left(\dfrac{110+90}{729}\right)=\dfrac{100}{729}$

Hence, $a+b=100+729=\boxed{829}$

$\begin{aligned} S & = \sum_{n=1}^\infty \frac{n^2+n}{2 \times 10^n} \\ & = \sum_{n=1}^\infty \frac{1}{10^{n}} \times \color{#3D99F6}{\frac{n(n+1)}{2}} \\ & = \sum_{n=1}^\infty \frac{\small{\color{#3D99F6}{ \displaystyle \sum_{k=1}^n k}}}{10^{n}} \\ & = \color{#3D99F6}{1} \sum_{n=\color{#D61F06}{1}}^\infty \frac{1}{10^n} + \color{#3D99F6}{2} \sum_{n=\color{#D61F06}{2}}^\infty \frac{1}{10^n} + \color{#3D99F6}{3} \sum_{n=\color{#D61F06}{3}}^\infty \frac{1}{10^n} + \color{#3D99F6}{4} \sum_{n=\color{#D61F06}{4}}^\infty \frac{1}{10^n} + ... \\ & =\frac{\color{#3D99F6}{1}}{\color{#D61F06}{10}} \sum_{n=\color{#D61F06}{0}}^\infty \frac{1}{10^n} + \frac{\color{#3D99F6}{2}}{\color{#D61F06}{10^2}} \sum_{n=\color{#D61F06}{0}}^\infty \frac{1}{10^n} + \frac{\color{#3D99F6}{3}}{\color{#D61F06}{10^3}} \sum_{n=\color{#D61F06}{0}}^\infty \frac{1}{10^n} + \frac{\color{#3D99F6}{4}}{\color{#D61F06}{10^4}} \sum_{n=\color{#D61F06}{0}}^\infty \frac{1}{10^n} + ... \\ & = \frac{1}{1-\frac{1}{10}} \sum_{n=1}^\infty \frac{n}{10^{n}} \\ & = \frac{10}{9} \times \frac{1}{10} \sum_{n=1}^\infty \frac{n}{10^{n-1}} \\ & = \frac{1}{9} \sum_{n=1}^\infty \frac{d x^n}{dx} \quad \quad \small \color{#3D99F6}{x = \frac{1}{10}} \\ & = \frac{1}{9} \times \frac{d}{dx} \sum_{n=1}^\infty x^n \\ & = \frac{1}{9} \times \frac{d}{dx} \left(\frac{x}{1-x}\right) \\ & = \frac{1}{9} \times \frac{1}{(1-x)^2} \quad \quad \small \color{#3D99F6}{\text{Putting back }x = \frac{1}{10}} \\ & = \frac{100}{729} \end{aligned}$

$\Rightarrow a + b = 100 + 729 = \boxed{829}$