Find the limit of the recursive sequence Definition  of  the  sequence :
$$a_1=a;\\a_2=b;\\$$ and $$\ \ \ a_{n+2}={{a_n+a_{n+1}}\over2}$$ for $n\geq 1$.
Find the limit of this  sequence in terms of $a$ and $b$.
Now in this case , taking $\lim_{n\rightarrow \infty}$  on  both sides does not help at all . So what I did was try to find the $n$-th term  in therms of $a$ and $b$ and then take $n$ to $\infty.$ 
So , using the given recursion formula I wrote down like  $7$-$8$  terms, grouped them  in  various  ways  but  still  could  not  get the pattern of the $n$-th term. 
Terribly  sorry  for  the  lack  of  work/context  in this post of mine but  I tried  for like an hour  to figure  things out  but can't really put  those  scribbles  down  here.
Please  give  me  some  hints  as  to  how  to find the answer .  Thank you.  
 A: Different solution:
\begin{align}
2a_{n+2}&=a_{n+1}+a_n\\
2(a_{n+2}-a_{n+1})&=-\left(a_{n+1}-a_n\right)\\
a_n-a_{n-1}&=-\frac12\left(a_{n-1}-a_{n-2}\right)\\
&=\frac14\left(a_{n-2}-a_{n-3}\right)=\cdots=\left(-\frac12\right)^{n-2}\left(a_2-a_1\right)\\
\therefore a_n&=a_{n-1}+\left(-\frac12\right)^{n-2}\left(b-a\right)\\
a_{n-1}&=a_{n-2}+\left(-\frac12\right)^{n-3}\left(b-a\right)\\
&\cdots\\
a_2&=a_1+\left(-\frac12\right)^{0}\left(b-a\right)\\
\end{align}
Adding side by side,
\begin{align}
a_n&=a_1+(b-a)\sum_{k=0}^{n-2}\left(-\frac12\right)^{k}\\
&=a+(b-a)\frac{1-\left(-\frac12\right)^{n-1}}{1-\left(-\frac12\right)}\\
&=a+\frac23(b-a)\left(1-\left(-\frac12\right)^{n-1}\right)\\
&=\frac{a+2b}{3}+\frac{2(a-b)}{3}\left(-\frac12\right)^{n-1}
\end{align}
You can see the outcome is exactly same as other solutions.
But hopefully you'll be able to find where the $\left(-\frac12\right)^{n-1}$ is coming from.
A: $$a_n=\frac{1}{3}a+\frac{2}{3}b+\left(\frac{4}{3}b-\frac{4}{3}a\right)\times\left(-\frac{1}{2}\right)^{n}$$
A: \begin{align}
&2a_{n+2}-a_{n+1}-a_n=0\\
&2\alpha^2-\alpha-1=0\quad\rightarrow\quad\alpha=1,-\frac12\\
&a_n=A\cdot1^{n-1}+B\left(-\frac12\right)^{n-1}\\
\end{align}
Using initial conditions,
\begin{align}
&A+B=a,\quad A-\frac B2=b\\
&\rightarrow A=\frac{a+2b}3,\quad B=\frac{2(a-b)}3\\
&a_n=\frac{a+2b}3+\frac{2(a-b)}3\left(-\frac12\right)^{n-1}\\
&\lim_{n\to\infty}a_n=\frac{a+2b}3\\
\end{align}
A: If you would like to have a closed form for the individual terms of the recurrence, you can use the method of generating functions. This first step places your sequence as the terms of a power series $G(x) = \sum_{n=1}^\infty a_n x^n$, and then you generate a separate representation using the recursion. It's a long process but very powerful. It's real strength comes from combinatorics, but it can solve linear recurrences well.
$$G(x) = \sum_{n=1}^\infty a_n x^n = ax + bx^2 + \sum_{n=3}^\infty a_n x^n$$
$$= ax + bx^2 + \sum_{n=1}^\infty a_{n+2} x^{n+2}$$
$$= ax + bx^2 + \frac x2 \sum_{n=1}^\infty a_{n+1} x^{n+1} + \frac{x^2}2 \sum_{n=1}^\infty a_{n} x^{n}$$
$$= ax + bx^2 + \frac x2 \left(\sum_{n=1}^\infty a_{n} x^{n} - ax\right) + \frac{x^2}2 \sum_{n=1}^\infty a_{n} x^{n}.$$
Thus recognizing $G(x)$ as the series on the right, we have
$$G(x)= ax + bx^2 + \frac x2 \left(G(x) - ax\right) + \frac{x^2}2 G(x).$$ Thus solving for $G(x)$ we find:
$$G(x)(1-\frac{x}{2}-\frac{x^2}{2})=ax + (b-\frac{a}{2})x^2$$
and
$$G(x) = \frac{2ax}{2-x-x^2} + \frac{(2b-a)x^2}{2-x-x^2}.$$
Using partial fraction decomposition we get:
$$G(x) = (2ax + (2b-a)x^2) \left( \frac{1}{6} \frac{1}{1-\left(-\frac x 2\right)} + \frac13 \frac{1}{1-x} \right)$$
Now applying geometric series identities:
$$G(x) = (2ax + (2b-a)x^2) \sum_{n=0}^\infty\left( \frac{1}{6}  \frac{(-1)^n}{2^n} + \frac13\right) x^n $$
Manipulating series:
$$G(x) = \sum_{n=0}^\infty 2a \left( \frac{1}{6}  \frac{(-1)^n}{2^n} + \frac13\right) x^{n+1} + \sum_{n=0}^\infty (2b-a)\left( \frac{1}{6}  \frac{(-1)^n}{2^n} + \frac13\right) x^{n+2}$$
$$=\sum_{n=1}^\infty 2a \left( \frac{1}{6}  \frac{(-1)^{n-1}}{2^{n-1}} + \frac13\right) x^{n} + \sum_{n=2}^\infty (2b-a)\left( \frac{1}{6}  \frac{(-1)^{n-2}}{2^{n-2}} + \frac13\right) x^{n}$$
$$=a + \sum_{n=2}^\infty \left[ 2a \left( \frac{1}{6}  \frac{(-1)^{n-1}}{2^{n-1}} + \frac13\right) + (2b-a)\left( \frac{1}{6}  \frac{(-1)^{n-2}}{2^{n-2}} + \frac13\right) \right] x^n$$
Thus since $G$ is a power series centered about zero, its coefficients are uniquely determined. For $n \ge 2$ we have: $$a_n = \left[ 2a \left( \frac{1}{6}  \frac{(-1)^{n-1}}{2^{n-1}} + \frac13\right) + (2b-a)\left( \frac{1}{6}  \frac{(-1)^{n-2}}{2^{n-2}} + \frac13\right) \right]$$
now as $n \to \infty$ we are left with $L = \lim_{n\to\infty} a_n = 2a \cdot \frac13 + (2b-a) \cdot \frac13 = \frac{1}{3}a + \frac{2}{3} b$.
A: We will try a form for the solution with some unknown parameters, and then see if we can find parameters that make it work. For example, let us guess that
$$
a_n = Ad^n
$$
for some constant numbers $d$ and $A$. Then we also need:
$$
a_{n+1} = Ad^{(n+1)}
$$
$$
a_{n+2} = Ad^{(n+2)}
$$
Then plug into our equation:
$$
Ad^{(n+2)} = \frac{Ad^{(n+1)}+Ad^n}{2}
$$
Every term is divisible by $Ad^n$, so cancelling that everywhere we get
$$
d^2 = \frac{d+1}{2}
$$
We can see that the recurrence relation is true if we can find a $d$ that satisfies this quadratic equation. Rearranging, we see that it is
$$
d^2-\frac{d}{2}-\frac{1}{2} = 0
$$
This quadratic factors nicely into
$$
(d-1)(d+1/2) = 0
$$
which is true when $d=1$ or $d=-1/2$. Now what this tells us is that if we take
$$
a_n = A1^n = A
$$
or
$$
a_n = A(-1/2)^n,
$$
we have a solution to the original problem. Recall that $A$ was an arbitrary constant that has not been determined yet, so it is possible that it is actually two different constants between the two cases above. To account for this, we write the second constant as $B$, giving two independent solutions
$$
a_n = A,\;\text{or}\;a_n=B(-1/2)^n
$$
Now, we can see that plugging in either solution solves the recurrence. What about their sum? You can work out that the sum of the two solutions is also a solution, so consider
$$
a_n = A + B(-1/2)^n
$$
This is the general solution. We figured out what the $d$ value was, now we have to figure out the $A$ and $B$ constants. Fortunately, we have some additional constraints:
$$
a_1 = a,\;a_2=b
$$
You should be able to take it from here to solve for $A$ and $B$, following what the other answers show.
A: Generalizing Kay K's solution:
If
$a_{n+1}
=ca_n+(1-c)a_{n-1}
$
where
$0 < c < 1$,
then
$a_{n+1}-a_n
=-(1-c)(a_n-a_{n-1})
=r(a_n-a_{n-1})
$,
where
$r = -(1-c)
$.
By induction,
$a_{n+1}-a_n
=r^k(a_{n-k+1}-a_{n-k-2})
$.
Setting
$k = n-1$,
$a_{n+1}-a_n
=r^{n-1}(a_{2}-a_{1})
=sr^{n-1}
$,
where
$s
=a_{2}-a_{1}
$.
Summing this from
$n=1$ to $m-1$,
$\begin{array}\\
a_m-a_1
&=s\sum_{n=1}^{m-1} r^{n-1}\\
&=s\sum_{n=0}^{m-2} r^{n}\\
&=s\frac{1-r^{m-1}}{1-r}\\
&=(a_{2}-a_{1})\frac{1-(c-1)^{m-1}}{c}\\
&=(a_{2}-a_{1})\frac1{c}-\frac{(c-1)^{m-1}}{c}\\
\end{array}
$
so
$a_m
=a_1+(a_{2}-a_{1})\frac1{c}-\frac{(c-1)^{m-1}}{c}
=\frac{a_{2}+(1-c)a_{1}}{c}-\frac{(c-1)^{m-1}}{c}
$.
Therefore,
since
$|c-1| < 1$,
$\lim_{m \to \infty}a_m
=\frac{a_{2}+(1-c)a_{1}}{c}$.
If $c = \frac12$,
this is
$\frac{a_{2}+\frac12 a_{1}}{\frac12}
=2a_{2}+a_{1}
$.
Note that
the characteristic polynomial
is
$0
=x^2-cx-(1-c)
=(x-1)(x-(c-1))
$.
This shows that
the roots
are always nice.
