Arithmetic-Geometric Series Application.

Let $|x| < 1$ .

$f(x) = \sum_{n = 0}^{\infty} (a + nd)x^n = a\sum_{n = 0}^{\infty} x^n + d\sum_{n = 1}^{\infty} nx^n =$ $\dfrac{a}{1 - x} + d\sum_{j = 1}^{\infty}\sum_{n = j}^{\infty} x^n =$ $\dfrac{a}{1 - x} + \dfrac{d}{1 - x}\sum_{j = 1}^{\infty} x^{j} = \dfrac{a}{1 - x} + \dfrac{d}{1 - x}(\dfrac{x}{1 - x}) = \boxed{\dfrac{a}{1 - x} + \dfrac{dx}{(1 - x)^2}}$

and,

$g(x) =\sum_{n = 0}^{\infty} (a^{*} + n^2 d^{*})x^n = a^{*}\sum_{n = 0}^{\infty} x^n + d^{*}\sum_{n = 1}^{\infty} n^2 x^n = \dfrac{a^{*}}{1 - x} + d^{*}\sum_{j = 1}^{\infty}\sum_{n = j}^{\infty} n x^n =$ $(\dfrac{a^{*}}{1 - x} + d^{*}\sum_{j = 1}^{\infty} (j x^{j} + (j + 1)x^{j + 1} + (j + 2)x^{j + 2} + ... ) =$ $\dfrac{a^{*}}{1 - x} + d^{*}(\dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j x^{j} + \dfrac{x}{(1 - x)^2}\sum_{j = 1}^{\infty} x^{j}) =$ $\dfrac{a^{*}}{1 - x} + d^{*}((\dfrac{1}{1 - x})(\dfrac{x}{(1 - x)^2}) + (\dfrac{x}{(1 - x)^2})(\dfrac{x}{1 - x})) = \boxed{\dfrac{a^{*}}{1 - x} + d^{*}(\dfrac{x^2 + x}{(1 - x)^3})}$

$f(\dfrac{1}{2}) = 2$ and $f(-\dfrac{1}{2}) = \dfrac{2}{9} \implies$

$3a - d = 1$

$a + d = 1$

$\implies a = \dfrac{1}{2} = d \implies f(x) = \dfrac{1}{2}(\dfrac{1}{(1 - x)^2}) \implies \displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) dx = \dfrac{1}{2}(\dfrac{1}{1 - x})|_{-\frac{1}{2}}^{\frac{1}{2}} = \boxed{\dfrac{2}{3}}$

$g(\dfrac{1}{2}) = 2$ and $g(-\dfrac{1}{2}) = \dfrac{2}{9} \implies$

$9a^{*} - d^{*} = 3$

$a^{*} + 3d^{*} = 1$

$\implies a^{*} = \dfrac{5}{14}$ and $d^{*} = \dfrac{3}{14}$ $\implies g(x) = \dfrac{1}{14}(\dfrac{5}{1 - x} + \dfrac{3(x^2 + x)}{(1 - x)^3})$

Let $I(x) = \displaystyle\int g(x) dx$ .

Let $u = 1 - x \implies x = 1 - u$ and $dx = -du$ $\implies I(x) = \dfrac{1}{14}(\displaystyle\int \dfrac{5}{1 - x} dx - 3\int (\dfrac{1}{u} - 3u^{-2} + 2u^{-3}) du) = \dfrac{1}{14}(-5\ln(1 - x) - 3(\ln(u) + \dfrac{3}{u} - \dfrac{1}{u^2})) =$ $\dfrac{1}{14}(-5\ln(1 - x) - 3\ln(1 - x) - \dfrac{9}{1 - x} + \dfrac{3}{(1 - x)^2}) = \dfrac{1}{14}(-8\ln(1 - x) + \dfrac{9x - 6}{(1 - x)^2}) \implies$

$\displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} g(x) dx = \boxed{\dfrac{1}{14}(8\ln(3) - \dfrac{12}{9})}$

$\implies \displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) - g(x) dx = \dfrac{4}{7}(\dfrac{4}{3} - \ln(3)) =$ $\dfrac{2^2}{7}(\dfrac{2^2}{3} - \ln(3)) = \dfrac{{\alpha}^{\alpha}}{\beta}(\dfrac{{\alpha}^{\alpha}}{\lambda} - \ln(\lambda)) \implies \alpha + \beta + \lambda = \boxed{12}$ .