Solving the recurrence $a_{n+1}-2a_n=2^n; a_0 = 1$ I am trying to solve the following recurrence but I am not sure which method is best for solving the following :
$$a_{n+1}-2a_n=2^n; a_0 = 1$$
My problem is that I know that this is not a linear equation and $2^n$ is not a polynomial equation so I'm not sure where to begin.
Any suggestions?
 A: Dividing the equation by $2^{n+1}$ we get
$$
\frac{a_{n+1}}{2^{n+1}}-\frac{a^n}{2^n}=\frac12
$$
So the sequence $b_n$ given by $b_n=\frac{a_n}{2^n}$ satisfies
$$
b_{n+1}-b_n=\frac12
$$
which is a lot easier to solve.
A: $$a_{n+1}-2a_n = 2(a_n-2a_{n-1})\Longrightarrow a_{n+1}-4a_n +4a_{n-1}=0$$
so the characteristic equation is $x^2-4x+4=0$ so $x_1=x_2=2$ so the general solution is $$a_n = (a\cdot n+b)\cdot 2^n$$
for some $a$ and $b$. Since $a_0=1$ we get $ b=1$ and since $a_1= 3 = 2(a+b)$ we get $a= 1/2$, so $$a_n = (n+2)\cdot 2^{n-1}$$
A: $$a_{n+1}-2a_n=2^n\Longleftrightarrow\frac{a_{n+1}}{2^{n+1}}-\frac{a_{n}}{2^{n}}=\frac{1}{2};$$
By telescopic sum we get
$$\frac{a_{n+1}}{2^{n+1}}-\frac{a_{0}}{2^{0}} =\sum_{k=0}^{n}\frac{a_{k+1}}{2^{k+1}}-\frac{a_{k}}{2^{k}}=\sum_{k=0}^{n}\frac{1}{2} =\frac{n+1}{2}$$ 
That is $$\color{red}{a_n =2^{n}(1+\frac{n}{2})= 2^{n-1}(n+2)}$$
A: Use the method of generating functions. Let $A(x) =\sum_{n=0}^\infty a_nx^n$. We wish to solve the recurrence relation
$$
a_{n+1}-2a_n=2^n; \quad a_0 = 1\tag{1}
$$
Mutliply both sides of (1) by $x^n$ and sum on $n$ to get that
$$
\frac{A(x)-1}{x}-2A(x)=\frac{1}{1-2x}.\tag{2}
$$
Solve for $A(x)$ to get that
$$
A(x)=\frac{1}{1-2x}+\frac{x}{(1-2x)^2}=\sum_{n=0}^\infty 2^nx^n+\sum_{n=1}^\infty n2^{n-1}x^n=\sum_{n=0}^\infty (2^n+n2^{n-1})x^n\tag{3}
$$
Hence
$$
a_n=2^n+n2^{n-1}=2^{n-1}(2+n).\tag{4}
$$
A: Note that $F(n)=(2)^n$. We need to keep in mind the following theorem.

Theorem: Suppose that $\{a_n\}$ satisfies the linear non-homogenous recurrence relation $$a_n=c_1a_{n-1}+c_2a_{n-2}+… c_ka_{n-k} +F(n)$$ where $c_1,c_2,…c_k \in \mathbb{R}$ and $F(n)= (b_tn^t+b_{t-1}n^{t-1}+…b_1n+b_0)s^n$, where $b_0, b_1, …b_t, s \in \mathbb{R}$.
When $s$ is not a root of the characteristic equation of the associated linear homogenous recurrence relation, there is a particular solution of the form $$(p_tn^t + p_{t-1}n^{t-1}+…p_1n+p_0)s^n$$
When $s$ is a root with multiplicity $m$, then the solution is of the form $$n^m(p_tn^t + p_{t-1}n^{t-1}+…p_1n+p_0)s^n$$

Taken from Rosen, Discrete Mathematics & Its Applications.
Coming back to the question, there is thus a particular solution of the form $n(p_1n+p_0)=p_1n^2+p_0n$ as $s=2$ is a root of degree $1$ of the characteristic equation ($x-2=0$).
Hope you can take it from here.
