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

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