Let an be sequence of real numbers with a1a21 and an25an16an

Let a_n be sequence of real numbers with a₁=a₂=1 and a_n+2=5a_n+1-6a_n for n=1,2,...

3, Let (an) be sequence of real numbers with a-a2 1 and Qn+i - 6a for n= 1,2, . i. Find the equation and its roots. ii. What is the general term of (an)?

Solution

a(n+2) = 5a(n+1) - 6an

a(n+2) is represented by r^2
a(n+1) by r
And an by constant term...

So, we get :

r^2 = 5r - 6

r^2 - 5r + 6 = 0

(r - 2)(r - 3) = 0

r = 2 and 3

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Now, general form is :

a(n) = A(2)^n + B(3)^n

We are given a1 = 1.....
So, 1 = A(2)^1 + B(3)^1
2A + 3B = 1 ----> euqation 1

We are given a2 = 1 :
So, 1 = A(2)^2 + B(3)^2
4A + 9B = 1 ---> equation 2

Equating, we get :

2A + 3B = 4A + 9B
2A = -6B
A = -3B

So, from equation 1 :
2A + 5B = 1
-6B+ 5B = 1
-B = 1
So, B = -1

And since A = -3B ,
A = -3(-1)
A = 3

So, the general form is :
a(n) = A(2)^n + B(3)^n

which now becomes .....

a(n) = (3).2^n - (1).3^n

So, we get :

a(n) = 3(2)^n - (3)^n

So, answers are :

Char eqn : r^2 - 5r + 6 = 0
Roots of char eqn : r = 2 , 3
general term an is : a(n) = 3(2)^n - (3)^n