3 3 votes 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 $\mathrm{G}(x)?$ Combinatory goclasses-scholarship-test1 numerical-answers goclasses combinatory generating-functions two-marks + – GO Classes 1.2k views answer comment Share Follow Print See all 2 Comments 2 2 Comments reply ankitgupta.1729 commented Aug 7, 2022 reply Follow flag $G(x) = \sum_{n=0}^{\infty}2^nx^n \times \sum_{r=0}^{\infty}f_rx^r$ $= \sum_{n=0}^{\infty} \sum_{r=0}^{\infty}\ 2^nf_r \ x^{n+r}$ $n+r=5 \implies n=5-r,$ So, $[x^5] \sum_{r=0}^{5}\ 2^{5-r}f_r $ Answer= $2^5.0+2^4.1+2^3.1+2^2.2+2^1.3+2^0.5 =43$ 2 2 replyShare Mr_Soprano commented Jul 20, 2025 reply Follow flag same question in today's scholarship test too :) 1 1 replyShare Please log in or register to add a comment.
3 3 votes $$\begin{aligned}&\text{G}(x)=\text{B}(x) \times \text{F}(x) \\&=\left(2^{0} x^{0}+2^{1} x^{1}+2^{2} x^{2}+2^{3} x^{3}+2^{4} x^{4}+2^{5} x^{5}+2^{6} x^{6}+\ldots\right) \times \\&\left(1 x^{1}+1 x^{2}+2 x^{3}+3 x^{4}+5 x^{5}+8 x^{6}+\ldots\right)\end{aligned}$$So, Coefficient of $x^{5}$ will be$:5+6+8+8+16=43.$Detailed Video Solution:https://youtu.be/tqjuxfutFHg?t=4221 GO Classes answered Aug 6, 2022 • edited Jul 20, 2025 by GO Classes GO Classes comment Share Follow 0 reply Please log in or register to add a comment.
