TITLE: If $a_n = \frac{1}{2} (a_{n-1}+a_{n-2})$ prove $a_n=2+4 \left(\frac{-1}{2}\right)^n$ for $n \geq 2$
QUESTION [7 upvotes]: Question: Let $a_1=0, a_2=3$, and, for all $n \geq 3$ let $a_n = \frac{1}{2} (a_{n-1}+a_{n-2})$ 
The sequence $(a_n)$ is said to be defined recursively.
on $n$, show that, for all $n \geq 2$,
$$a_n=2+4 \left(\frac{-1}{2}\right)^n$$
And deduce that $(a_n) \rightarrow 2$


My attempt:
$
  a_1 = 0 , a_2 = 3 , \ldots
$
$\forall n \geq 3$ let $a_n = \frac{1}{2}(a_{n-1} + a_{n-2})$
Show inductively that $\forall n \geq 2$
$
  a_n = 2 + 4(- \frac{1}{2})^n
$
And deduce that $(a_n) \rightarrow 2$
So $\ldots$
let $n$ = $3$ then $a_3 = \frac{1}{2}(a_2 + a_{1}) = \frac{1}{2}(3 + 0) = \frac{3}{2}$
This is equal to $2 + 4(- \frac{1}{2})^3$, so this holds.
Assume that this holds up to $n = k$, then for $n = k + 1$;
$a_{k + 1} = \frac{1}{2}(a_k + a_{k-1})$
$a_k$ is known as $a_k = 2 + 4(- \frac{1}{2})^k$
And $a_{k - 1}$ is also known as $a_{k - 1}= 2 + 4(- \frac{1}{2})^{k - 1}$
so I then have
$2a_{k+1} = (a_k + a_{k-1})$, and subbing in the values above for $a_k$ and
$a_{k-1}$ gives
$2a_{k+1} = ((2 + 4(- \frac{1}{2})^k) + (2 + 4(- \frac{1}{2})^{k - 1}))$
Which can be simplified to
$a_{k+1} = (1 + 2(- \frac{1}{2})^k + 1 + 2(- \frac{1}{2})^{k - 1})$
and
$a_{k+1} = 2 + 2(- \frac{1}{2})^k + 2(- \frac{1}{2})^{k - 1}$
Using exponent laws for $ 2(- \frac{1}{2})^{k - 1}$ gives
\begin{equation*}
  \begin{aligned}
    2(- \frac{1}{2})^{k - 1} &= 2(-\frac{1}{2})^{k} \times (- \frac{1}{2})^{-1} \\
    &= 2(-\frac{1}{2})^{k} \times (-2) \\
    &= -4(-\frac{1}{2})^{k} \\
  \end{aligned}
\end{equation*}
Subbing this back into $a_{k+1} = 2 + 2(- \frac{1}{2})^k + 2(- \frac{1}{2})^{k -
  1}$  as
\begin{equation*}
  \begin{aligned}
    a_{k+1} &= 2 + 2(- \frac{1}{2})^k -4(-\frac{1}{2})^{k} \\
    &= 2 - 2(- \frac{1}{2})^k
  \end{aligned}
\end{equation*}
If the induction holds then
$ a_{k + 1} =  2 - 2(- \frac{1}{2})^k = 2 + 4(- \frac{1}{2})^{k + 1}$
Using power laws on $ 4(- \frac{1}{2})^{k + 1}$ gives
\begin{equation*}
  \begin{aligned}
    4(- \frac{1}{2})^{k + 1} &= 4(- \frac{1}{2})^{k } \times (- \frac{1}{2})^{1} \\
    &= 4(- \frac{1}{2})^{k } \times (- \frac{1}{2}) \\
    &= -2(- \frac{1}{2})^{k }  \\
  \end{aligned}
\end{equation*}
Using this result gives
\begin{equation*}
  \begin{aligned}
    a_{k + 1} &=  2 - 2(- \frac{1}{2})^k \\
    &= 2 + 4(- \frac{1}{2})^{k + 1} \\
    &= 2  -2(- \frac{1}{2})^{k }  \\
  \end{aligned}
\end{equation*}
As required.
Therefore by the inductive hypothesis $\forall n \geq 2, a_n = 2 + 4(- \frac{1}{2})^n $
Deduce that $(a_n) \rightarrow 2$
If $(a_n) \rightarrow 2$ then as $n \rightarrow \infty, a_n \rightarrow 2$.
So the limit of the sequence is $2$.
Using the definition of convergence:
$\forall \epsilon > 0$ there exist natural numbers $N,n$ where $n>N$ such that
$|2 + 4(- \frac{1}{2})^n - 2| < \epsilon$
Which simplifies to $|4(- \frac{1}{2})^n| < \epsilon$ and $4(\frac{1}{2})^n <
\epsilon$
Then, given $\epsilon > 0$ choose $N$ such that the minimum value of $N$ is
$\log_2(\frac{4}{\epsilon})$ such that for all $n > N, a_n < \epsilon$
This can be seen as
\begin{equation*}
  \begin{aligned}
    N &> \log_2(\frac{4}{\epsilon}) \\
    2^N &> \frac{4}{\epsilon} \\
    \frac{1}{2^N}  &< \frac{\epsilon}{4} \\
    \frac{4}{2^N}  &< \epsilon \\
  \end{aligned}
\end{equation*}
As required.
This proves that $a_n \rightarrow 2$
Not sure if this is correct but could someone please check and if there is an easier way of proving this let me know!

REPLY [5 votes]: By a well known theorem the general solution of $a_n = \dfrac{1}{2}(a_{n-1} + a_{n-2})$ can be found by means of the characteristic equation:
$$\lambda^2=\dfrac{1}{2}(\lambda+1)=0\Leftrightarrow \ldots \Leftrightarrow \lambda=1\vee\lambda=-1/2,\text{ so }a_n=C_1(1)^n+C_2\left(-\dfrac{1}{2}\right)^n.$$ $$n=1\Rightarrow 0=C_1-\dfrac{1}{2}C_2,\quad n=2\Rightarrow 3=C_1+\dfrac{1}{4}C_2.$$ Solving the system we get $C_1=2,\;C_2=4$, so $$\boxed{\;a_n=2+4\left(-\dfrac{1}{2}\right)^n\;}$$