The solution of the recurrence relation an = 10a_n-1 - 25a_n-2,  is:
(a₀ = 1, a₁ = 15)

Concept:

We use the characteristics roots method for repeated roots to solve the above recurrence relation.

If we have an recurrence relation as an + c1an-1 + c2an-2 = 0, then the characteristics equation is given as x2 + c1x + c2 = 0 .

If r is the repeated root of the characteristics equation then the solution to recurrence relation is given as \(a_n=ar^n+bnr^n\) where a and b are constants determined by initial conditions.

Calculation:

The recurrence relation is an = 10an-1 - 25an-2  with initial conditions a0 = 1, a1 = 2.

The recurrence relation can be written as  an = 10an-1 - 25an-2 = 0  and the characteristics equation is given as 

x2 - 10x + 25 = 0.

Solve for x,

(x - 5) (x - 5) = 0

x = 5 or x = 5 

So, r = 5 is the repeated root of the characteristics equation. So, the solution of recurrence relation is given as

an = a5n + bn5n

an = (a + bn)5n       ......(i)

No, we will find a and b with the given initial values a0 = 1, a1 = 15.

For a0 = 1,   

1 = (a + b.0)50   

a = 1       ......(ii)

For, a1 = 15

15 = (a + b.1)51

5(a + b) = 15

a + b = 3       ......(iii)

Solving equation (ii) and (iii), we get

a = 1 and b = 2.

Putting these values in equation (i) we get,

an = (1 + 2n)5n

Hence, the solution of given recurrence relation is an = (1 + 2n)5n