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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?

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7 Answers 7

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 $a_n= 2 + 2^{1-n}$

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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 $n \ge 1$

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Your solution doesn't satisfy a2 or a1 –  Ishan Banerjee Jan 24 '13 at 10:42
    
@IshanBanerjee $n \ge 1$. Put $n = 1$ and you generate all values. –  hjpotter92 Jan 24 '13 at 10:44
    
Putting n=1, we get 3/2 instead of 5/2. Your solution is off by one for all values of n. –  Ishan Banerjee Jan 24 '13 at 10:49
    
@IshanBanerjee Thanks. :) I'd summed up the GP to n terms, instead of $n-1 \dots$ :| –  hjpotter92 Jan 24 '13 at 10:51
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Note that $$\begin{align}a_{n+1} &=1+\frac{a_n}{2} =1+\frac{1}{2}(1+\frac{a_{n-1}}{2})\\ &=1+\frac{1}{2}+\frac{1}{2^2}a_{n-1} \\ &=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}a_{n-2}\\ & \vdots \\ &= 1+\frac{1}{2}+\frac{1}{2^2}+ \dots+\frac{1}{2^{n-1}}+\frac{1}{2^n}a_1 \\ &=1+\frac{1}{2}+\frac{1}{2^2}+ \dots+\frac{1}{2^{n-1}}+\frac{3}{2^n}\end{align}$$

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$a_2=1+3/2=5/2=(2^2+1)/2^1$

$a_3=1+5/4=9/4=(2^3+1)/2^2$

$a_4=1+9/8=17/8=(2^4+1)/2^3$

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$a_n=(2^n+1)/2^{n-1}$

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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

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Actually f can't be determined from this information. Only the value of f at integer points can be determined from this. –  Ishan Banerjee Jan 24 '13 at 11:36
    
Check out linear recurrence relations on Wikipedia. It gives a variety of methods to solve something like this. –  Ishan Banerjee Jan 24 '13 at 13:31
    
I am Sorry Here $x\in \mathbb{Z}$.have you a link or document about solving these type of Question. Thanks –  juantheron Jan 24 '13 at 15:20
    
en.wikipedia.org/wiki/… –  Ishan Banerjee Jan 24 '13 at 15:35
    
Ask questions as questions, not in answers to other questions. You should probably delete this answer, and ask again (taking the comments here into consideration). –  vonbrand Jan 24 '13 at 19:16
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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, $$ a_n = 2^{1-n}+2.$$

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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: $$ a_n = \begin{cases} 3 & n = 1 \\ 5 \cdot 2^{n - 1} & n > 1 \end{cases} $$

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