Answer to Question #197186 in Discrete Mathematics for sakibur rahman

Question #197186

Find the solution of the recurrence relation: xn=3xn-1 + 1, where,  x0=4


1
Expert's answer
2021-05-24T16:21:54-0400

Let us find the solution of the recurrence relation: xn=3xn1+1x_n=3x_{n-1} + 1, where x0=4.x_0=4.

For this firstly, let us solve the characteristic equation of the homogeneous equation xn3xn1=0x_n-3x_{n-1}=0 :

k3=0k-3=0 or k=3.k=3. It follows that the particular solution of the equation is xn=a=constx_n=a=const, where a=3a+1,a=3a+1, and hence a=12.a=-\frac{1}{2}. The solution of the recurrence relation: xn=3xn1+1x_n=3x_{n-1} + 1

is xn=C3n12.x_n=C\cdot3^n-\frac{1}{2}. Taking into account that x0=4,x_0=4, we conclude that 4=x0=C124=x_0=C-\frac{1}{2}, and thus C=92.C=\frac{9}{2}. Consequently,

xn=923n12=3n+212.x_n=\frac{9}{2}\cdot3^n-\frac{1}{2}=\frac{3^{n+2}-1}{2}.


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