Question $4(1\mathrm{.}5$ points$)$

The Fibonacci sequence $1,1,2,3,5,8,13,21,34,55,$dots $is$ usually defined

recursively via

$c_0=1,c_1=c_0,c_n=c_{n-1}+c_{n-2}$ for $n2.T(z)={\displaystyle \underset{n=0}{\overset{}{}}}{c}_n{z}^n=1+z+2z^2+3z^3+5z^4+8z^5+13z^6+21z$

$--$ what happens?

Select $4$ correct answerfs$(1+z+z^2)*T(z)=1,so$ wherever the series converges, $itis$ identical $to$

$f(z)=\frac{1}{1+z+z^2}$

Multiplying out and comparing coefficients shows that

$(1-z-z^2)*T(z)=1,so$ wherever the series converges, $itis$ identical $to$

$f(z)=\frac{1}{1-z-z^2}=\frac{-1}{z^2+z-1}.$

Multiplying out and comparing coefficients shows that

$(1-z+z^2)*T(z)=1,so$ wherever the series converges, $itis$ identical $to$

$f(z)=\frac{1}{1-z+z^2}.$

The function $f$ identified above has two $singularities,at$ the zeros $of$ its

denominator. The Maclaurin Series $off$ thus converges $on$ the disk around $0$

whose radius $is$ the smaller $of$ the $absolute$ values $of$ these roots: This number $is$

$\sqrt[\phantom{2}]{5}-1$~~$1\mathrm{.}23.$

The function $f$ identified above has two $singularities,at$ the zeros $of$ its

denominator. The Maclaurin Series $off$ thus converges $on$ the disk around $0$

whose radius $is$ the smaller $of$ the