Find the 101st term!

Consider the generating function:

$\quad \quad f(x)={ a }_{ 0 }+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+{ a }_{ 3 }{ x }^{ 3 }+...\\ 2xf(x)=\quad \quad 2{ a }_{ 0 }x+{ 2a }_{ 1 }{ x }^{ 2 }+{ 2a }_{ 2 }{ x }^{ 3 }+2{ a }_{ 3 }{ x }^{ 4 }+...\\ 3{ x }^{ 2 }f(x)=\quad \quad \quad \quad \quad \quad 3{ a }_{ 0 }{ x }^{ 2 }+{ 3a }_{ 1 }{ x }^{ 3 }+{ 3a }_{ 2 }{ x }^{ 4 }+{ a }_{ 3 }{ x }^{ 5 }+...\\ \Rightarrow f(x)(1-2x-3{ x }^{ 2 })={ a }_{ 0 }+({ a }_{ 1 }-2{ a }_{ 0 })x+\sum _{ i=2 }^{ \infty }{ { (3x) }^{ i } } \\ \Rightarrow f(x)(1-2x-3{ x }^{ 2 })=-2+\sum _{ i=0 }^{ \infty }{ { (3x) }^{ i } } =-2+\frac { 1 }{ 1-3x } \\ \Rightarrow f(x)=\frac { 2 }{ (3x-1)(x+1) } +\frac { 1 }{ (x+1){ (1-3x) }^{ 2 } }$

Now we use partial fractions to get:

$f(x)=\frac { 3 }{ 4{ (1-3x) }^{ 2 } } +\frac { 21 }{ 16(3x-1) } -\frac { 7 }{ 16(x+1) }$

Using binomial theorem for negative indices , we can now find the closed form formula for $a_n$ .

Hence, $a_{100}=\frac { 397({ 3 }^{ 101 })-7 }{ 16 }$

Therefore the answer is $397+3+101+7+16=\boxed{524}$

Multiply the recurrence relation by $x^n$ and afterwards eliminate all negative index-shifts with $n':=n-2,\quad n'\rightarrow n$ : $\begin{aligned} && a_{n+2}x^{n+2} &= 2x\cdot a_{n+1}x^{n+1}+3x^2\cdot a_nx^n + 9x^2\cdot (3x)^n &&&&&&\left| \sum_{n=0}^\infty \ldots\right. \quad \left|A(x) :=\sum_{n=0}^\infty a_nx^n\right.\\[1.5em] \Rightarrow&& A(x)-a_0-a_1x &= 2x\left( A(x) -a_0 \right) + 3x^2A(x) + \frac{9x^2}{1-3x} &&&&&& \left| \red{(*)}\text{ for }|x|<\frac{1}{3} \right. \end{aligned}$

Insert the initial conditions $(a_0,\:a_1)=(-1;\:1)$ and solve for $A(x)$ . Simplify the result with partial fraction decomposition (PFD): $\begin{aligned} A(x) &=\frac{\cancel{\mp 9x^2} + 6x-1}{(1-3x)(1-2x-3x^2)} = \frac{6x-1}{(1-3x)^2(1+x)}\underset{\text{PFD}}{=}\frac{3}{4}\cdot\frac{1}{(1-3x)^2} - \frac{21}{16}\cdot\frac{1}{1-3x} -\frac{7}{16}\cdot\frac{1}{1+x}\\[1.5em] &\underset{\red{(*)}}{=}\sum_{n=0}^\infty \left[ \frac{3}{4}\cdot\binom{n+1}{n}3^n - \frac{21}{16}\cdot 3^n - \frac{7}{16}\cdot (-1)^n \right]x^n\overset{!}{=}\sum_{n=0}^\infty a_nx^n,\quad |x|<\frac{1}{3} \end{aligned}$

Comparing coefficients we can reconstruct the sequence $a_n$ : $\begin{aligned} a_n&=\frac{3}{4}\cdot\binom{n+1}{n}3^n - \frac{21}{16}\cdot 3^n - \frac{7}{16}\cdot (-1)^n=\frac{(4n-3)3^{n+1} - 7(-1)^n}{16} &&&\Rightarrow &&&& a_{100} &=\frac{397\cdot 3^{101}-7}{4^2} \end{aligned}$

The answer $v+w+x+y+z=4^2 + 7 + 397+3+101=\boxed{524}$ satisfies all conditions for $v,\:w,\:x,\:y,\:z$ .

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$\begin{aligned} \text{generalized geometric sum:} &&&& \sum_{n=0}^\infty\binom{n+m}{n}q^n &= \frac{1}{(1-q)^{m+1}} & \text{for} && |q| &< 1, & m & \in\mathbb{N}_0 & \red{(*)} \end{aligned}$

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Rem.: If we introduce $z:=x^{-1}\in\mathbb{C}$ , the generating function $A(x)$ becomes $A(z^{-1})$ , the $\mathcal{Z}$ -transform of $a_k$ from digital signal theory!