Limit of a recursive sequence Let $\lambda$$\in$$(0,1)$. For any real $a_0$, $a_1$, define the sequence recursively by $$a_n = (1-\lambda)a_{n-1}  + \lambda a_{n-2}$$ 
Let $\alpha$ = $\lim\limits_{n\rightarrow\infty}a_n$
Express the $\alpha$ in terms of $\lambda$, $a_0$ and $a_1$
 A: Look at this. For your recurrence, the characteristic equation is $$m^2 - (1- \lambda)m - \lambda = 0$$ Rearranging we get $$m^2 - m + \lambda (m-1) = m(m-1) + \lambda (m-1) = (m+\lambda)(m-1)= 0$$ This gives us $m = 1$ (or) $m = \lambda$. The solution is then given by $a_n = c_1 + c_2 (-\lambda)^n$. Since $\lambda \in (0,1)$, $\lim_{n \rightarrow \infty} a_n = c_1$. Hence, we need to find $c_1$. Setting $n=0$, we get $a_0 = c_1 + c_2$ and $a_1 = c_1 - \lambda c_2$. Solving for $c_1$, we get $\displaystyle c_1 = \frac{\lambda a_0 + a_1}{1+ \lambda}$. Hence, $$\lim_{n \rightarrow \infty} a_n = \frac{\lambda a_0 + a_1}{1+ \lambda}.$$
A: It is important to become acquainted with the characteristic equation method. 
However, since this problem seems to come from an analysis course, and not a combinatorics course, we will solve it without using the characteristic equation machinery.
We first describe some geometric preliminaries, though in fact none of the geometry is used in the solution!  But it is always good to see what is really going on. 
Geometric preliminaries:   If $p$ and $q$ are numbers, or indeed points in the plane or in space, and $0<\lambda<1$, then $\lambda p +(1-\lambda)q$ is the point on the ray that goes from $p$ to $q$, and divides that ray in the ratio $\lambda:1-\lambda$. In particular, $\lambda p +(1-\lambda)q$ is between $p$ and $q$ if $0 <\lambda<1$. The case $a_0=a_1$ is uninteresting, since then $a_n=a_0$ for all $n$. 
Draw a picture. Without loss of generality we may assume that  $a_0<a_1$.  A little playing will show that the sequence $a_0, a_2, a_4, \dots$ is increasing, and the sequence $a_1,a_3,a_5,\dots$ is decreasing. Also, $a_{2k}<a_{2k+1}$. 
The sequence $(a_{2k})$ is increasing and bounded above, so has a limit. Similarly, the sequence $(a_{2k+1})$ is bounded below, and therefore has a limit. We show that the gap between successive $a_i$ goes to $0$ has $i$ goes to infinity. That will show that the limits of $(a_{2k})$ and $(a_{2k+1})$ are the same.  
The equation
$$a_n=(1-\lambda)a_{n-1}+\lambda a_{n-2}$$
can be rewritten as
$$a_n-a_{n-1}=-\lambda(a_{n-1}-a_{n-2}). \qquad\text{(Equation 1)}$$
Let $c=a_1-a_0$.
Then, by Equation $1$, $a_2-a_1=-\lambda(a_1-a_0)=c(-\lambda)$. Therefore, by Equation $1$, $a_3-a_2=-\lambda(a_2-a_1)=c(-\lambda)^2$.
But $a_4-a_3=-\lambda(a_3-a_2)=c(-\lambda)^3$.  And from that we obtain $a_5-a_4=c(-\lambda)^4$.  The pattern is obvious. We have
$$a_{n}-a_{n-1}=c(-\lambda)^{n-1}. \qquad\text{(Equation 2)} $$
As $n \to \infty$, $\lambda^{n-1}\to 0$, so the gap between $a_{n-1}$ and $a_n$ goes to $0$.  It follows that the limits of $(a_{2k})$ and $(a_{2k+1})$ are the same, so we know now that $\lim_{n\to\infty}a_n$ exists.
The solution: We will use none of material in the geometric preliminaries, except for Equation $1$, which is not really geometric.
We can rewrite Equation $1$ as
$$a_n=a_{n-1}+c(-\lambda)^{n-1}.$$
So $a_n$ is obtained by adding $c(-\lambda)^{n-1}$ to $a_{n-1}$. A familiar procedure, the one that produces a geometric series! 
We have $a_1=a_0+ c$, and therefore $a_2=a_0+c+c(-\lambda)$, and therefore $a_3=a_0+c +c(-\lambda)+c(-\lambda)^2 $, and so on. In general, for $n \ge 1$, 
$$a_n=a_0+ c+c(-\lambda) +c(-\lambda)^2 + c(-\lambda)^3 +\cdots +c(-\lambda)^{n-1}.$$
Apart from the term $a_0$ at the front, we can see that $a_n$ is the partial sum of a geometric series with first term $c$ and common ratio $-\lambda$. 
We can now use the formula for the sum of a finite geometric series to get a simple formula for $a_n$, and then take the limit. 
Or else we can see that by the definition of the sum of an infinite geometric series, we have
$$\lim_{n\to \infty} a_n=a_0+c\sum_{n=0}^\infty (-\lambda)^n.$$
When $|\lambda| <1$, the inner infinite series converges, and has sum $\frac{1}{1-(-\lambda)}$. It follows that 
$$\lim_{n\to \infty} a_n=a_0+\frac{a_1-a_0}{1+\lambda}.$$ 
Comment: We describe an idea that would have made the proof much more attractive.  The geometry of the situation does not change if we subtract $a_0$ from $a_0$ and $a_1$.  The geometry of the situation also does not really change if we scale the interval from $0$ to $a_1-a_0$ by the factor $1/(a_1-a_0)$. Formally, let
$$b_k=\frac{a_k -a_0}{a_1-a_0}.$$
Then $b_0=0$, $b_1=1$, and the sequence $(b_n)$ satisfies the same recurrence as the sequence $(a_n)$.
The formulas become more attractive. We find that 
$$b_n=1-\lambda+\lambda^2-\lambda^3+ \cdots +(-1)^{n-1}\lambda^{n-1}=\frac{1-(-\lambda)^n}{1+\lambda},$$
so the sequence $(b_n)$ has limit $\frac{1}{1+\lambda}$.
Now use the fact that $b_n=\frac{a_n -a_0}{a_1-a_0}$ to find the limit of the sequence $(a_n)$.
A: I just want to give an easy answer to this question.
So, It is given , $\alpha=\lim_{n\to\infty} a_n \implies \lim_{n\to\infty} a_{n-1}=\alpha$            (*)
And , $a_n=(1-\lambda)a_{n-1}+\lambda a_{n-2}$ , we can write it as $a_n+\lambda a_{n-1}=a_{n-1} +\lambda a_{n-2}$ $\forall n\ge 2$
Now, let  $ A_n=a_n+\lambda a_{n-1}$ and $A_{n-1}=a_{n-1} +\lambda a_{n-2}$ .
Note that $A_n=A_{n-1}=A_{n-2}=...=A_3=A_2$
$\implies$  $a_n+\lambda a_{n-1}=a_1+\lambda a_0$ 
After taking limit , we'll get $\alpha+\lambda \alpha=a_1+\lambda a_0$ (from *)
Hence , $\alpha=\frac{a_1+\lambda a_0}{1+\lambda}$ .
