Solve the recurrence relation√a_{n}=√a_{n-1}+ 2√a_{n-2}with initial conditions a_{0}= a_{1}= 1 by making the substitution b_{n}=√a_{n}.

I made the substitution from the beginning. Never seen an example like this. My working...

b_{n}= 1b_{n-1}+ 2b_{n-2}

Characteristic equation form: x^{2}= ax + b implies x^{2}= 1x + 2

x^{2}-x -2 = 0

(x - 2)(x + 1) = 0

x = 2, -1.

General solution is b_{n}= A(2)^{n}+ B(-1)^{n}

When n = 0; b_{0}=√a_{0}=√1 = 1.

1 = A(2)^{0}+ B(-1)^{0}

1 = A(1) + B

When n = 1; b_{1}=√a_{1}=√1 = 1.

1 = A(2)^{1}+ B(-1)^{1}

1 = 2A - B

Solving simultaneous equations...

A = 2 / 3.

B = 1 / 3.

Solution is (2/3)2^{n}+ (1/3)(-1)^{n}.

Method is correct?