Find the generating function of the sequence $\{a_{n}\}$ where $a_{n}=2n+3$ for all $n=0,1,2,\dots$
Generating sequence$: a_{0} = 3 , a_{1} = 5, a_{2} = 7,a_{3} = 9, a_{4} = 11,\dots$
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 $\langle \:3,5,7,9,11,\dots\rangle$
We can write the generating function:
$G(x) = 3 + 5x +7x^{2} + 9x^{3} + 11x^{4}+\dots \rightarrow(1)$
Lets generating sequence are $\langle\: 1,1,1,1,1,1,1,1,1,\dots \rangle$
We can find the generating function:
$1+ x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6} +\dots \Leftrightarrow \dfrac{1}{(1-x)} \rightarrow (2)\:\:\:\: [\because\text{Sum of Infinite series}]$
Multiply $'2x'$ both sides,we get
$2x+ 2x^{2} + 2x^{3} + 2x^{4} + 2x^{5} +2x^{6} + 2x^{7} +\dots \Leftrightarrow \dfrac{2x}{(1-x)}$
Differentiate both side with respect to $'x'$ and get,
$\dfrac{\mathrm{d} }{\mathrm{d} x}(2x+ 2x^{2} + 2x^{3} + 2x^{4} + 2x^{5} +2x^{6} + 2x^{7} +\dots)\Leftrightarrow \dfrac{\mathrm{d} }{\mathrm{d} x}\left(\dfrac{2x}{1-x}\right)$
$2+ 4x + 6x^{2} + 8x^{3} + 10x^{4} +\dots \Leftrightarrow \dfrac{(1-x).2-2x(-1)}{(1-x)^{2}}$
$2+ 4x + 6x^{2} + 8x^{3} + 10x^{4} +\dots \Leftrightarrow \dfrac{2-2x+2x}{(1-x)^{2}}$
$2+ 4x + 6x^{2} + 8x^{3} + 10x^{4} +\dots \Leftrightarrow \dfrac{2}{(1-x)^{2}} \rightarrow(3)$
Adding the equation $(2)$ and equation $(3),$we get,
$1+ x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6} +\dots \Leftrightarrow \dfrac{1}{(1-x)}$
$2+ 4x + 6x^{2} + 8x^{3} + 10x^{4} +\dots \Leftrightarrow \dfrac{2}{(1-x)^{2}}$
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$3+ 5x + 7x^{2} + 9x^{3} + 11x^{4} +\dots \Leftrightarrow \dfrac{1}{(1-x)} + \dfrac{2}{(1-x)^{2}}$
$G(x) = \dfrac{1}{(1-x)} + \dfrac{2}{(1-x)^{2}}$
$G(x) = \dfrac{(1-x) + 2}{(1-x)^{2}}$
$G(x)=\dfrac{3-x }{(1-x)^{2}}$
$$\textbf{(OR)}$$
From the equation $(2),$
$1+ x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6} +\dots \Leftrightarrow \dfrac{1}{(1-x)} \rightarrow (2)\:\:\:\: [\because\text{Sum of Infinite series}]$
Differentiate both side with respect to $'x'$ and get,
$1+ 2x + 3x^{2} + 4x^{3} + 5x^{4} + 6x^{5} + 7x^{6} +\dots \Leftrightarrow \dfrac{1}{(1-x)^{2}}\rightarrow (3)\:\:\: \left[\because \left(\dfrac{u}{v}\right)^{'} = \dfrac{u'\:v - u\:v'}{v^{2}}\right]$(quotient rule)
In equations $(3)$ multiply $'3'$ both sides, and we get
$3+ 6x + 9x^{2} + 12x^{3} + 15x^{4} + 18x^{5} + 21x^{6} +\dots \Leftrightarrow \dfrac{3}{(1-x)^{2}}\rightarrow (4)$
In equations $(3)$ multiply $'x'$ both sides, and we get
$3x+ 6x^{2} + 9x^{3} + 12x^{4} + 15x^{5} + 18x^{6} +\dots \Leftrightarrow \dfrac{x}{(1-x)^{2}}\rightarrow (5)$
Subtract the equation $(4)$ and equation $(5),$we get,
$3+ 6x + 9x^{2} + 12x^{3} + 15x^{4} + 18x^{5} + 21x^{6} +\dots \Leftrightarrow \dfrac{3}{(1-x)^{2}}\rightarrow (4)$
$3x+ 6x^{2} + 9x^{3} + 12x^{4} + 15x^{5} + 18x^{6} +\dots \Leftrightarrow \dfrac{x}{(1-x)^{2}}\rightarrow (5)$
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$3+5x+7x^{2} +9x^{3}+11x^{4} + 13x^{5} + 15x^{6} + \dots \Leftrightarrow \dfrac{3}{(1-x)^{2}} - \dfrac{x}{(1-x)^{2}}$
$3+5x+7x^{2} +9x^{3}+11x^{4} + 13x^{5} + 15x^{6} + \dots \Leftrightarrow \dfrac{3-x}{(1-x)^{2}}$
So, the correct answer is $(D).$