Find the generating function of the sequence $\{a_{n}\}$ where $a_{n}=\dfrac{1}{2}n(n + 3)$ for all $n=1,2,3,\dots$
Generating sequence$: a_{0} = 2 , a_{1} = 5, a_{2} = 9,a_{3} = 14, a_{4} = 20,\dots$
We know that Ordinary Generating Function, given that generating sequence are $\langle a_{0},a_{1},a_{2},a_{3},a_{4},\dots\rangle$
$$G(x) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + a_{4}x^{4}+\dots$$
$$G(x) = \sum_{k=0}^{\infty} a_{k}\cdot x^{k}$$
$\therefore$ In question given sequence is $\langle \:2,5,9,14,20\dots\rangle$
We can write the generating function:
$G(x) = 2 + 5x +9x^{2} + 14x^{3} + 14x^{4}+20x^{5} + \dots \rightarrow(1)$
Lets generating sequence are $\langle\: 1,1,1,1,1,1,1,1,1,\dots \rangle$
We can find the generating function:
$1+ x + x^{2} + x^{3} + x^{4} + x^{5} + \dots \Leftrightarrow \dfrac{1}{(1-x)} \rightarrow (2)\:\:\:\: [\because\text{Sum of Infinite series}]$
Differentiate both side with respect to $'x'$ and get,
$1+ 2x + 3x^{2} + 4x^{3} + 5x^{4} + 6x^{5} +\dots \Leftrightarrow \dfrac{1}{(1-x)^{2}}\rightarrow (3)\:\:\: \left[\because \left(\dfrac{u}{v}\right)^{'} = \dfrac{u'\:v - u\:v'}{v^{2}}\right]$(quotient rule)
$$\textbf{Not Complete Yet}$$