13.10. Suppose that a sequence (ao, a1, a2, ) of real numbers satisfies the recurrence relation an -5an-1+6an-20 for all n> 2. (a) What is the order of the linear recurrence relation? (b) Express the generating function of the sequence as a rational function. (c) Find a generic closed form solution for this recurrence relation. (d) Find the terms ao,a1,.. . ,a5 of this sequence when the initial conditions are given by ao 2 and a5 (e) Find the closed form solution when ao 2 and a 5.

a. This recurrence is of order 2.b. We're looking for a function [tex]A(x)[/tex] such that[tex]A(x)=\displaystyle\sum_{n=0}^\infty a_nx^n[/tex]Take the recurrence,[tex]\begin{cases}a_0=a_0\\a_1=a_1\\a_n-5a_{n-1}+6a_{n-2}=0&\text{for }n\ge2\end{cases}[/tex]Multiply both sides by [tex]x^{n-2}[/tex] and sum over all integers [tex]n\ge2[/tex]:[tex]\displaystyle\sum_{n=2}^\infty a_nx^{n-2}-5\sum_{n=2}^\infty a_{n-1}x^{n-2}+6\sum_{n=2}^\infty a_{n-2}x^{n-2}=0[/tex]Pull out powers of [tex]x[/tex] so that each summand takes the form [tex]a_kx^k[/tex]:[tex]\displaystyle\frac1{x^2}\sum_{n=2}^\infty a_nx^n-\frac5x\sum_{n=2}^\infty a_{n-1}x^{n-1}+6\sum_{n=2}^\infty a_{n-2}x^{n-2}=0[/tex]Now shift the indices and add/subtract terms as needed to get everything in terms of [tex]A(x)[/tex]:[tex]\displaystyle\frac1{x^2}\left(\sum_{n=0}^\infty a_nx^n-a_0-a_1x\right)-\frac5x\left(\sum_{n=0}^\infty a_nx^n-a_0\right)+6\sum_{n=0}^\infty a_nx^n=0[/tex][tex]\displaystyle\frac{A(x)-a_0-a_1x}{x^2}-\frac{5(A(x)-a_0)}x+6A(x)=0[/tex]Solve for [tex]A(x)[/tex]:[tex]A(x)=\dfrac{a_0+(a_1-5a_0)x}{1-5x+6x^2}\implies\boxed{A(x)=\dfrac{a_0+(a_1-5a_0)x}{(1-3x)(1-2x)}}[/tex]c. Splitting [tex]A(x)[/tex] into partial fractions gives[tex]A(x)=\dfrac{2a_0-a_1}{1-3x}+\dfrac{3a_0-a_1}{1-2x}[/tex]Recall that for [tex]|x|<1[/tex], we have[tex]\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n[/tex]so that for [tex]|3x|<1[/tex] and [tex]|2x|<1[/tex], or simply [tex]|x|<\dfrac13[/tex], we have[tex]A(x)=\displaystyle\sum_{n=0}^\infty\bigg((2a_0-a_1)3^n+(3a_0-a_1)2^n\bigg)x^n[/tex]which means the solution to the recurrence is[tex]\boxed{a_n=(2a_0-a_1)3^n+(3a_0-a_1)2^n}[/tex]d. I guess you mean [tex]a_0=2[/tex] and [tex]a_1=5[/tex], in which case[tex]\boxed{\begin{cases}a_0=2\\a_1=5\\a_2=13\\a_3=35\\a_4=97\\a_5=275\end{cases}}[/tex]e. We already know the general solution in terms of [tex]a_0[/tex] and [tex]a_1[/tex], so just plug them in:[tex]\boxed{a_n=2^n+3^n}[/tex]