Partial sum formula for $n^{2}(-3)^{-n}$

Question

I was solving a linear recurrence given by: $$a_{n}+3a_{n-1}=4n^{2}-2n+2^{n}$$ $$a_{0}=1$$ Here's what I've done: $$(-3)^{n}b_{n}:=a_{n}$$ Substituting in the first equation: $$b_{n}-b_{n-1}=4n^{2}(-3)^{-n}-2n(-3)^{-n}+\left(-\frac{2}{3}\right)^{n}$$ The trouble is to find a telescoping sum for the first term of the formula, which consists essentially of: $$n^{2}(-3)^{-n}$$ I am able to find the second and the third terms. Thanks in advance.

Answer 1

To solve the linear recurrence, it is advisable to decompose the RHS as a sum of zero (homogeneous equation), a polynomial and an exponential, and solve separately.

1. $$a_n+3a_{n-1}=0\iff a_n=(-3)a_{n-1}\iff a_n=(-3)^na_0.$$

2. $$a_n+3a_{n-1}=4n^2-2n$$ is solved with a quadratic poynomial, such that the leading term is $n^2$, because $n^2+3(n-1)^2=4n^2+\cdots$.

Let $a_n=n^2+an+b$ and

$$n^2+an+b+3(n-1)^2+3a(n-1)+3b=4n^2-2n.$$

By identification, one finds

$$a_n=n^2+n.$$

3. $$a_n+3a_{n-1}=2^n$$ is solved with the same exponential, $c2^n$,

$$a_n+3a_{n-1}=c2^n+3c2^{n-1}=2^n\iff c=\frac25.$$