Solution to Solve recurring relation :an+2 -10an+1+25an=5n(n≥0) - Sikademy
Author Image

Archangel Macsika

Solve recurring relation :an+2 -10an+1+25an=5n(n≥0)

The Answer to the Question
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

Let us solve the recurrence relation a_{n+2} -10a_{n+1}+25a_n=5^n,\ (n≥0). The characteristic equation k^2-10k+25=0 of homogeneous relation is equivalent to (k-5)^2=0, and hence it has the roots k_1=k_2=5. Therefore, the solution of the non-homogeneous recurrence relation is a_n=(C_1+C_2n)5^n+p_n, where the particular solution of non-homogeneous recurrence relation is p_n=an^25^n.


It follows that

a(n+2)^25^{n+2}-10a(n+1)^25^{n+1}+25an^25^n=5^n, which is equivalent to

25a(n^2+4n+4)5^{n}-50a(n^2+2n+1)5^{n}+25an^25^n=5^n, and hence

25an^2+100an+100a-50an^2-100an-50a+25an^2=1.

We conclude that 50a=1, and thus a=\frac{1}{50}.


Consequently, the solution of the recurrence relation a_{n+2} -10a_{n+1}+25a_n=5^n is

a_n=(C_1+C_2n)5^n+\frac{1}{50}n^25^n or a_n=(C_1+C_2n)5^n+\frac{1}{2}n^25^{n-2}.


Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-5-stid-8-sqid-2755-qpid-1225