Iterated sequence - general formula I have a question regarding writing a general formula for an iterated sequence of variables. 
Say we have 
$$a_0=1$$ 
$$a_1=\frac{3}{2}a_0$$ 
$$a_2=\frac{3}{2}a_1-\frac{1}{2}a_0$$ 
$$a_3=\frac{3}{2}a_2-\frac{1}{2}a_1$$
So in general we have $a_i=\frac{3}{2}a_{i-1}-\frac{1}{2}a_{i-2}$. The solution this this iterated sequence is $a_i=2-2^{-i}$. What method should one use to come up with this solution? ? 
Thanks in advance. 

Starting from your general term 
$$a_n=\frac{3}{2}a_{n-1}-\frac{1}{2}a_{n-2}$$
substituting value of $a_{n-1}$ in this , it will transform into $$a_{n}=\frac{7}{4}a_{n-2}-\frac{3}{4}a_{n-3}$$
which is also equals to $$a_n=\frac{2^{2+1}-1}{2^2}a_{n-2}-\frac{2^2-1}{2^2}a_{n-3}$$
again substituting the value of $a_{n-2}$ in it , it will transform into (I hope you will be understanding from where I am substituting values, of course from your general terms)$$a_n=\frac{15}{8}a_{n-3}-\frac{7}{8}a_{n-4}$$
which is also equals to$$a_n=\frac{2^4-1}{2^3}a_{n-3}-\frac{2^3-1}{2^3}a_{n-4}$$
conclusion goes to$$a_n=\frac{2^{x+1}-1}{2^x}a_{n-x}-\frac{2^x-1}{2^x}a_{n-(x+1)}$$
put $x=n-1$ and take it from here, you will get your original answer that is $$a_n=2-2^{-n}$$
