How to solve $$a_n=4a_{n-1}-4a_{n-2}$$ $a_0=6$ and $a_1=8$
I found first few terms as $$6,8,8,0,... $$ But I don't know how to proceed.
Asked
Active
Viewed 158 times
-1
-
1Another one: https://math.stackexchange.com/q/1544784/42969. – Martin R Nov 13 '18 at 06:21
-
1Many many many questions about homogeneous linear recurrences have been asked and answered here before. Maybe do a little search for something like yours. – Gerry Myerson Nov 13 '18 at 06:21
-
Yes...I should have looked that earlier. Thanks anyway – Nov 13 '18 at 06:28
1 Answers
0
Make its characteristic equation: $$x^2-4x+4=0$$
Solving which gives repeated root as $x=2$
So, its solution will be like: $$a_n=(A+Bn)2^n$$
Just substitute for initial conditions and you get $A=6$ and $B=-2$
So, $$a_n=(6-2n)\cdot2^n$$
pooja somani
- 2,575