Find the generating function for Ar and use it to evaluate Ar...

$\left( \sum_{k = 2}^{\infty} C(k, 2)x^k \right)(1 + x + x^2 + x^3 +...)$

$= [C(2,2)]x^2 + [C(3,2) + C(2,2)]x^3 + [C(4, 2) + C(3,2) + C(2,2)]x^4 + ...$

$\text{so } \left( \sum_{k = 2}^{\infty} C(k, 2)x^k \right)\left( \frac{1}{1-x} \right) = \sum_{k = 2}^{\infty} A_{k-2}x^k.$

$\text{Now }\sum_{k = 2}^{\infty} C(k, 2)x^k = x^2\sum_{k = 0}^{\infty} C(k+2, 2)x^k$

$\text{and }C(k+2, 2)x^k = \frac{(k+2)!}{2! \ k!}x^k = \frac{1}{2} (k+2)(k+1)x^k = \frac{1}{2} \frac{d^2}{dx^2}(x^{k+2})$

$\text{so }\sum_{k = 2}^{\infty} C(k, 2)x^k = x^2\sum_{k = 0}^{\infty} \frac{1}{2} \frac{d^2}{dx^2}(x^{k+2})$

$= \frac{x^2}{2} \frac{d^2}{dx^2} \left(\sum_{k = 0}^{\infty} x^{k+2} \right)$

$= \frac{x^2}{2} \frac{d^2}{dx^2} \left(x^2\sum_{k = 0}^{\infty} x^k \right)$

$= \frac{x^2}{2} \frac{d^2}{dx^2} \left(\frac{x^2}{1-x} \right)$

$= \frac{x^2}{2} \frac{d}{dx} \left(\frac{(2x)(1-x) - (x^2)(-1)}{(1-x)^2} \right)$

$= \frac{x^2}{2} \frac{d}{dx} \left(\frac{2x-x^2}{(1-x)^2} \right)$

$= \frac{x^2}{2} \left( \frac{(2-2x)(1-x)^2 - (2x-x^2)[2(1-x)(-1)]}{(1-x)^4} \right)$

$= \frac{x^2}{2} \left( \frac{2 \ [ \ (1-x)^2 + (2x-x^2) \ ]}{(1-x)^3} \right)$

$= \frac{x^2}{2} \left( \frac{2}{(1-x)^3} \right)$

$= \frac{x^2}{(1-x)^3}.$

$\text{Thus } \sum_{k = 2}^{\infty} A_{k-2}x^k = \left( \sum_{k = 2}^{\infty} C(k, 2)x^k \right)\left( \frac{1}{1-x} \right)$

$= \left( \frac{x^2}{(1-x)^3} \right)\left( \frac{1}{1-x} \right) = \frac{x^2}{(1-x)^4}.$

$\text{Since } \sum_{k = 2}^{\infty} A_{k-2}x^k = x^2 \sum_{k = 0}^{\infty} A_kx^k,$

$\text{have } x^2 \sum_{k = 0}^{\infty} A_kx^k = \frac{x^2}{(1-x)^4}.$

$\text{Therefore } \sum_{k = 0}^{\infty} A_kx^k = \frac{1}{(1-x)^4}.$

This the generating function for $A_k$ .

Now $A_k$ is calcuable by finding the Taylor series for that function about x=0. Its derivatives are very easy to compute and very simple to write down in general. I'll leave that to you.