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Let $\text{S}$ denote the infinite sum $2+5x+9x^{2}+14x^{3}+20x^{4}+\ldots$

where $\mid x \mid < 1$ and the coefficient of $x^{n-1}$ is $\dfrac{1}{2}n\left ( n+3 \right ), \left ( n=1,2,\ldots \right ).$ Then $\text{S}$ equals

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

0 votes

Find the generating function of the sequence $\{a_{n}\}$ where $a_{n}=\dfrac{1}{2}n(n + 3)$ f*or* 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)

$$\textbf{Not Complete Yet}$$