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Define the generating functions $\text{B}(x)=\displaystyle{} \sum_{n=0}^{\infty} 2^{n} x^{n}$ and $F(x)=\displaystyle{} \sum_{n=0}^{\infty} f_{n} x^{n}$ where $f_{n}$ is the Fibonacci sequence determined by the recurrence relation
\begin{aligned} f_{0} &=0 \quad \text { and } \quad f_{1}=1 \\ f_{n} &=f_{n-1}+f_{n-2} & \text { for } n \geq 2\\ \end{aligned}
Let $\text{G}(x)=\text{B}(x) \times \text{F}(x)$.

What is the coefficient of $x^{5}$ is $\mathrm{G}(x)?$

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$G(x) = \sum_{n=0}^{\infty}2^nx^n \times \sum_{r=0}^{\infty}f_rx^r$

$= \sum_{n=0}^{\infty} \sum_{r=0}^{\infty}\ 2^nf_r \ x^{n+r}$

$n+r=5 \implies n=5-r,$ So,

$[x^5] \sum_{r=0}^{5}\ 2^{5-r}f_r$

Answer= $2^5.0+2^4.1+2^3.1+2^2.2+2^1.3+2^0.5 =43$

\begin{aligned} &\text{G}(x)=\text{B}(x) \times \text{F}(x) \\ &=\left(2^{0} x^{0}+2^{1} x^{1}+2^{2} x^{2}+2^{3} x^{3}+2^{4} x^{4}+2^{5} x^{5}+2^{6} x^{6}+\ldots\right) \times \\ &\left(1 x^{1}+1 x^{2}+2 x^{3}+3 x^{4}+5 x^{5}+8 x^{6}+\ldots\right) \end{aligned}
So, Coefficient of $x^{5}$ will be$:5+6+8+8+16=43.$

Detailed Video Solution:

https://youtu.be/tqjuxfutFHg?t=4221