# Rosen 7e Exercise-8.2 Question no-23 page no-525 Recurrence Relation

89 views
Consider the nonhomogeneous linear recurrence relation $a_n$=$3a_{n-1}$+$2^n$

in the book solution is given $a_n$=$-2^{n+1}$

but I’m getting $a_n$=$3^{n+1}-2^{n+1}$

edited
0
what is the base condition ?
$-2^{n+1}$ is the particular solution..
1
base condition is $a_0$=1
im getting the same

homogeneous part i got $3^{n+1}$
and particular part $-2^{n+1}$
1

You can verify yourself by taking small values of 'n'.

For eg.,  for $n=1$, recurrence relation gives $a_1=3a_0+2^1=3*1+2=5$

Now,$-2^{n+1}$ will give answer as $-2^2=-4$ where as $3^{n+1}-2^{n+1}$ will give answer as $3^2-2^2=5$. So, your answer is correct.

## Related questions

1
134 views
What is the general form of the particular solution guaranteed to exist of the linear nonhomogeneous recurrence relation $a_n$=$6a_{n-1}$-$12a_{n-2}$+$8a_{n-3}$+F(n) if F(n)=$n^2$ F(n)=$2^n$ F(n)=$n2^n$ F(n)=$(-2)^n$ F(n)=$n^22^n$ F(n)=$n^3(-2)^n$ F(n)=3
Consider the nonhomogeneous linear recurrence relation $a_{n} = 3a_{n-1} + 2^{n}.$ Show that $a_{n} = -2^{n+1}$ is a solution of this recurrence relation. Use Theorem $5$ to find all solutions of this recurrence relation. Find the solution with $a_{0} = 1.$