Given generating function $G(z) = \dfrac{1 + z}{(1-z)^{3}}$
We know that if given sequence is
$(1,1,1,1,1,1,\dots) \Leftrightarrow \dfrac{1}{1-z}$
$\left[\text{Infinite series summation}\: S = \dfrac{a}{1-r}; r< 1\: \text{(or)}\: S = \dfrac{a}{r-1};r > 1\: \text{Where a = first term and r = common ratio} \right]$
$(1 + z + z^{2} + z^{3} + z^{4} + \dots) \Leftrightarrow \dfrac{1}{1-z}$
Differentiate both side with respect to $z$
$\dfrac{\mathrm{d} }{\mathrm{d} z} (1 + z + z^{2} + z^{3} + z^{4} + \dots) \Leftrightarrow \dfrac{\mathrm{d} }{\mathrm{d}z } \left(\dfrac{1}{1-z}\right)$
$(0+1+2z+3z^{2}+4z^{3}+\dots) \Leftrightarrow \dfrac{1}{(1-z)^{2}}$
$(1+2z+3z^{2}+4z^{3}+\dots) \Leftrightarrow \dfrac{1}{(1-z)^{2}}$
Multiply both side by $z$
$(z+2z^{2}+3z^{3}+4z^{4}+\dots) \Leftrightarrow \dfrac{z}{(1-z)^{2}}$
Again differentiate both side with respect to $z$
$(1+4z+9z^{2}+16z^{3}+\dots) \Leftrightarrow \dfrac{(1-z)^{2}(1) - z(2(1-z))(-1)}{((1-z)^{2})^{2}}$
$(1+4z+9z^{2}+16z^{3}+\dots) \Leftrightarrow \dfrac{1+z^{2}-2z +2z(1-z)}{(1-z)^{4}}$
$(1+4z+9z^{2}+16z^{3}+\dots) \Leftrightarrow\dfrac{1+z^{2}-2z +2z-2z^{2}}{(1-z)^{4}}$
$(1+4z+9z^{2}+16z^{3}+\dots) \Leftrightarrow \dfrac{1+z^{2}-2z^{2}}{(1-z)^{4}}$
$(1+4z+9z^{2}+16z^{3}+\dots) \Leftrightarrow \dfrac{1^{2}-z^{2}}{(1-z)^{4}}$
$(1+4z+9z^{2}+16z^{3}+\dots) \Leftrightarrow\dfrac{(1-z)(1+z)}{(1-z)^{4}}$
$(1+4z+9z^{2}+16z^{3}+\dots) \Leftrightarrow \dfrac{(1+z)}{(1-z)^{3}} = G(z)$
$G(z) = 1+4z+9z^{2}+16z^{3}+\dots\:\:\:\rightarrow(1)$
Ordinary Generating Function
$G(z) = a_{0} + a_{1}z +a_{2}z^{2} +a_{3}z^{3}+a_{4}z^{4}+\dots\:\:\:\rightarrow(2)$
Compare equation $(1)$ and equation $(2)$ and we get
$a_{0} =1 $
$a_{3}=16$
Now we can easily find $a_{3} -a_{0}=16-1=15$
So, the correct answer is $15.$
See this pdf for generating function it might be helpful.
see here
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$\textbf{Second Method:}$
- $1+x+x^{2}+x^{3}+\dots + x^{n} = \dfrac{1-x^{n+1}}{1-x}$
- $\dfrac{1}{(1-x)^{n}} = \displaystyle{}\sum^{\infty}_{0} \binom{n+k-1}{k}\:x^{k}$
Given generating function $G(z) = \dfrac{1 + z}{(1-z)^{3}}$
$\implies G(z) = (1+z)\:\displaystyle{}\sum^{\infty}_{0} \binom{3+k-1}{k}\:z^{k}$
$\implies G(z) = (1+z)\:\displaystyle{}\sum^{\infty}_{0} \binom{2+k}{k}\:z^{k}$
$\implies G(z) = (1+z)\left[1+3z+6z^{2} + 10z^{3} + \dots \right]$
$\implies G(z) = 1+3z+6z^{2} + 10z^{3} + \dots + z+3z^{2}+6z^{3} + 10z^{4} + \dots$
$\implies G(z) = 1+4z+9z^{2} + 16z^{3} + \dots$
Ordinary Generating Function
$G(z) = a_{0} + a_{1}z +a_{2}z^{2} +a_{3}z^{3}+a_{4}z^{4}+\dots\:\:\:\rightarrow(1)$
$a_{0} =1 ,a_{3}=16$
$ \therefore a_{3} -a_{0}=16-1=15$
So, the correct answer is $15.$