Question

9. Use advancement operators to find a closed-form formula for xn, as a 0, function of n, given that forn2 and that

applied combinatorics

Answer 1

Given reccurence relation is

$x_{n+3}=7x_{n+2}-16x_{n+1}+12x_n+2^n$___(*)

Since the given reccurence relation is non homogenuous ,

Guess

Now put all the values in (*).

n +3

                        $=7c2^{n+2}+7d-16(c2^{n+1}-16d+12c2^{n}+12d+2^n$

                         $=7c2^{n+2}-16c2^{n+1}+12c2^{n}+2^n+3d$

$7c2^{n+2}-16c2^{n+1}+12c2^{n}+2^n-c2^{n+3}+2d=0$

$7c2^{n+2}-16c2^{n+1}+(1+12c)2^{n}-c2^{n+3}+2d=0$

Now compare the coef ,

Hence

is the particular soolution .

For homogenuous ,

$h_{n+3}=7h_{n+2}-16h_{n+1}+12h_n$

$1=7r-16r^2+12r^3$

(2r-1 )2(3r-1)-0

$h_n=(A+Bn)(1/2)^n+C(1/3)^n$

Hence solution is

$x_n=y_n+h_n=-1/12(2)^n+(A+Bn)(1/2)^n+C(1/3)^n$

Intial conditions

$x_0=0=-1/12(2)^0+(A+B*0)(1/2)^0+C(1/3)^0$

                   

$x_1=0=-1/12(2)^1+(A+B*1)(1/2)^1+C(1/3)^1$

                   

xi = 0 =-1/6 + (A + B)/2 + C/3

$x_2=1=-1/12(2)^2+(A+B*2)(1/2)^2+C(1/3)^2$

                  $1=-1/3+(A+2B)/4+C/9$

$(9A+18B)+4C=12--(3)$

By (1),(2),(3)

$A=-20/3,B=5/2,C=27/4$

Hence

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