About the solution of a difference equation

Question

Let $r>4$ be a positive integer. Let us consider this difference equation: $$u_{n+1}=(1+r²ⁿ⁺¹)u_{n}-r²ⁿ⁻¹u_{n-1}+2$$

I want to find a closed form, bu I am not able to find the good idea.

Answer 1

HINT: $$u_{n+1}=(1+r^{2n+1})u_{n}-r^{2n-1}u_{n-1}+2$$ $$(u_{n+1}-r^{2n+1}u_n)=(u_n-r^{2n-1}u_{n-1})+2$$ Find the sum of first $n-1$ terms. Then we have a first order, first degree difference equation $$u_n-r^{2n-1}u_{n-1}=(u_1-ru_0)+2(n-1).$$ Note that $2n-1=n^2-(n-1)^2,$ therefore $$u_n-r^{2n-1}u_{n-1}=r^{n^2}\Big(\dfrac{u_{n}}{r^{n^2}}-\dfrac{u_{n-1}}{r^{(n-1‌​)^2}}\Big).$$ $$\dfrac{u_{n}}{r^{n^2}}-\dfrac{u_{n-1}}{r^{(n-1‌​)^2}}=\dfrac{(u_1-ru_0)+2(n-1)}{r^{n^2}}.$$ Then add first $n$ terms, then you will have a closed form for $u_n$ .