Given that the sequence x1 x2 is such that x1 0 x2 4 and x

Given that the sequence (x_1, x_2, ...) is such that x_1 = 0, x_2 4, and x_n + 1 = 4(x_n-1), prove that x_n = (n - 1)2^n. Find the general term of such a sequence if the first two terms are given instead as x_1 = 1 and x_2 = 3.

Solution

It is a linear homogeneous recurrence relation so we assume solution

x_n=r^n

Substituting gives

r^2=4r-4

r^2-4r+4=0

r=2, hence repeated roots

So general solution is

x_n=2^n(A+Bn)

x_1=2(A+B)=0 hence, A=-B

x_2=4(A+2B)=4 hence, A+2B=1

HEnce, B=1, A=-1

So general solution is

x_n=2^n(n-1)

Now using the second set of intial conditions we get

x_1=2(A+B)=1

Hence, A+B=1/2

x_2=4(A+2B)=3

A+2B=3/4

Hence, B=1/4,A=1/4

So general solution is

x_n=2^n(1+n)/4