Solve $a_n - 4a_{n-1} + 4a_{n-2} = 2^n$ Solve $a_n - 4a_{n-1} + 4a_{n-2} = 2^n$
given that $a_0 = 0$, and $a_1 = 3$
My Attempt:
Get the characteristic equation and solve it.
For homogeneous equation
$x^2 -4x + 4 = 0$
$x = 2  $ or $  x = 2$
Hence, $a_n^h = (A+Bn)\cdot2^n $
Guess a particular solution: $n^22^nC$
$n^22^nC - 4(n-1)^22^{n-1}C + 4(n-2)^22^{n-2}C = 2^n$
$n^2C - 2(n-1)^2C + (n-2)^2C = 1$
$C= \frac12$
Hence, $a^p_n = \frac12n^22^n$
$a_n = a^p_n + a^h_n$
$a_n = (A+Bn)\cdot2^n + \frac12n^22^n$
$a_0 = 0 = A$
$a_1 = 3 = 2B + 1$
$B = 1$
Therefore, $a_n = (n+ \frac12n^2)2^n$
 A: No, $n^22^{n^2}C$ does not work. Try with $n^22^{n}C$. Then
$$2^n=a_n - 4a_{n-1} + 4a_{n-2} = n^22^{n}C-4(n-1)^22^{n-1}C+4(n-2)^22^{n-2}C=2^{n}\cdot 2C$$
which implies that $C=1/2$.
In general, if the r.h.s. is $r^n$ and $r$ is a solution of the characteristic polynomial of multiplicity $m$, then a particular solution has the form $n^m C r^n$.
Note that $n^i r^n$ is a solution of the homogeneous recurrence for $0\leq i<m$.
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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$\ds{a_{n} - 4a_{n - 1} + 4a_{n - 2} = 2^{n}\,;\qquad
a_{0} = 0\,,\quad a_{1} = 3}$.

\begin{align}
&a_{n} - 4a_{n - 1} + 4a_{n - 2} = 2^{n} \implies
1 = {a_{n} \over 2^{n}} - 2\,{a_{n - 1} \over 2^{n - 1}} +
{a_{n - 2} \over 2^{n - 2} }
\end{align}

Moreover,
\begin{align}
\sum_{n = 2}^{\infty}z^{n} & =
\sum_{n = 2}^{\infty}{a_{n} \over 2^{n}}\,z^{n} -
2\sum_{n = 2}^{\infty}{a_{n - 1} \over 2^{n - 1}}\,z^{n} +
\sum_{n = 2}^{\infty}{a_{n - 2} \over 2^{n - 2}}\,z^{n}
\\[5mm]
{z^{2} \over 1 - z} & =
\sum_{n = 2}^{\infty}{a_{n} \over 2^{n}}\,z^{n} -
2\sum_{n = 1}^{\infty}{a_{n} \over 2^{n}}\,z^{n + 1} +
\sum_{n = 0}^{\infty}{a_{n} \over 2^{n}}\,z^{n + 2} 
\\[5mm] & =
\pars{\sum_{n = 0}^{\infty}{a_{n} \over 2^{n}}\,z^{n} - {3 \over 2}\,z} -
2z\sum_{n = 0}^{\infty}{a_{n} \over 2^{n}}\,z^{n} +
z^{2}\sum_{n = 0}^{\infty}{a_{n} \over 2^{n}}\,z^{n}
\\[5mm] & =
\pars{1 - z}^{2}\sum_{n = 0}^{\infty}{a_{n} \over 2^{n}}\,z^{n}  - {3 \over 2}\,z
\end{align}

Then,
\begin{align}
\sum_{n = 0}^{\infty}{a_{n} \over 2^{n}}\,z^{n} & =
-\,{z^{2} - 3z \over 2\pars{1 - z}^{3}} =
-\,{1 \over 2}\sum_{n = 0}^{\infty}{-3 \choose n}\pars{-1}^{n}
\pars{z^{n + 2} - 3z^{n + 1}}
\\[5mm] & =
-\,{1 \over 4}\sum_{n = 0}^{\infty}\pars{n + 2}\pars{n + 1}
\pars{z^{n + 2} - 3z^{n + 1}}
\\[5mm] & =
-\,{1 \over 4}\sum_{n = 2}^{\infty}n\pars{n - 1}z^{n} +
{3 \over 4}\sum_{n = 1}^{\infty}\pars{n + 1}n\,z^{n}
\\[5mm] & =
{3 \over 2}\,z +
\sum_{n = 2}^{\infty}
\bracks{-\,{1 \over 4}\,n\pars{n - 1} + {3 \over 4}\,\pars{n + 1}n}\,z^{n} =
{3 \over 2}\,z +
\sum_{n = 2}^{\infty}{\pars{n + 2}n \over 2}\,z^{n}
\\[5mm] & =
\sum_{n = 0}^{\infty}{\pars{n + 2}n \over 2}\,z^{n} \implies
{a_{n} \over 2^{n}} = {\pars{n + 2}n \over 2} \implies
\bbox[10px,#ffe,border:0.1em groove navy]{\color{#f00}{a_{n}}  =
\color{#f00}{\pars{n + 2}n\,2^{n - 1}}\,,\quad \forall\ n \geq 0}
\end{align}
