Find the nth term of a recursive sequence I have a the following sequence:
$$\begin{gather}
a_1 = 3 \\
a_{n + 1} = 1 + \frac{a_n}{2}
\end{gather}
$$
How can I find the $a_n$ term? 
 A: Let $a_n=b_n +2$
Then. $b_{n+1}=\frac{b_n}{2}$
and $b_1=1$
So, $b_n=2^{1-n}$
So $a_n= 2 + 2^{1-n}$
A: Note that $$\begin{align}a_{n+1} &=1+\frac{a_n}{2} =1+\frac{1}{2}(1+\frac{a_{n-1}}{2})\\ &=1+\frac{1}{2}+\frac{1}{2^2}a_{n-1} \\ &=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}a_{n-2}\\ &  \vdots \\ &= 1+\frac{1}{2}+\frac{1}{2^2}+ \dots+\frac{1}{2^{n-1}}+\frac{1}{2^n}a_1 \\ &=1+\frac{1}{2}+\frac{1}{2^2}+ \dots+\frac{1}{2^{n-1}}+\frac{3}{2^n}\end{align}$$
A: Put $n = 1$; you get $a_2 = 1 + \frac{1}{2} * 3 = \frac{5}{2}$
Following this continually, you get:
$$a_{n} = \frac{3 + \Sigma {2^{n - 1}}}{2^{n - 1}}$$
where, $n = 1, 2, 3 .. $
Aletnatively
You have $a_2 = 1 + \frac{a_1}{2}$
$$a_3 = 1 + \frac{a_2}{2} = 1 + \frac{1}{2} * (1 + \frac{a_1}{2}) = 1 + \frac{1}{2} + \frac{a_1}{4}$$
Hence, in this manner, you generate:
$$a_{n + 1} = 1 + \frac{1}{2} + ... \text{n terms} + \frac{a_1}{2^n} = \frac{1 * (1 - \frac{1}{2^n}) }{1 - \frac{1}{2}} + \frac{3}{2^n}$$
Therefore, you conclude:
$$a_{n + 1} = 2 + \frac{1}{2^n}$$
where $n \ge 1$
A: $a_2=1+3/2=5/2=(2^2+1)/2^1$
$a_3=1+5/4=9/4=(2^3+1)/2^2$
$a_4=1+9/8=17/8=(2^4+1)/2^3$
.
.
$a_n=(2^n+1)/2^{n-1}$
A: I have a Same Question
If $f(x+2)-3f(x+1)+2f(x) = 0\; \forall x\in \mathbb{R}$ and $f(1)=4$ and $f(2) = 6$ .Then $f(x) = $
Thanks
A: All solutions to the recurrence relation $a_{n+1} = s a_n +t $ with $s \neq 1$ have the form: 
$$ 
a_n= c_1 s^n +c_2,
$$ 
where $c_1$ and $c_2$ are specific constants.
In the problem $s= 1/2$. Therefore, $a_n= c_1 (1/2)^n + c_2 $. 
Taking into account $a_0=4$ and $a_1=3$, one can obtain $c_1=2=c_2$. Hence,
$$ a_n = 2^{1-n}+2.$$ 
A: Let's use generating functions a bit on this one. Define the ordinary generating function $A(z) = \sum_{n \ge 0} a_{n + 1} z^n$. From the recursion we have by properties of the ordinary generating function:
$$
\frac{A(z) - a_1}{z} = 1 + \frac{A(z)}{2}
$$
As $a_1 = 3$, this gives:
$$
A(z) = \frac{5}{1 - z / 2} - 2
$$
The first term is just a geometric series. This tells us that:
$$
a_n = \begin{cases}
         3 & n = 1 \\
         5 \cdot 2^{n - 1} & n > 1
      \end{cases}
$$
