Question

Solve the following recurrence relation an=2an-1+an-2-2an-3 with initial conditions a0=3 a1=6 a2=0.

Answer 1

Answer:

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Answer 2

The value of C1 is -1 and the value of C2 is 3

Given:

The recurrence relation of an=2an-1+an-2-2an-3 with initial conditions a0=3 a1=6 a2=0

To Find:

To get the value of C1 and C2 using the recurrence relation

Solution:

Let us find the solution of the recurrence relation,

an=2an-1+an-2-2an-3

Let us solve the characteristic equation

k^2=k+2

=k+2 which is equivalent to

k^2-k-2=0

and hence by Vieta's formulas has the solutions

k1=-1

and k2=2.k

follows that the solution of the equation is ,

an = C1$(-1)^{n} + C2 .2^n$

Since ,

a0 = 2

and a1 = 7

we have that ,

2 = $a_{0}$=C1 +C2  and

7 = -C1 +2C2

Therefore,

C2 =3 and C1 = -1

We could conclude that,

an = $(-1)^ n+1 + 3.2^n$

Hence, we get the equation as an = $(-1)^ n+1 + 3.2^n$

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