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I want to test the following using mathematical induction.

If $x_1=1$, $x_2=2$ and $x_{n+2}=\frac{1}{2}(x_n+x_{n+1})$ prove that $1 \leq x_n \leq 2$ for all $n$ natural

My idea was to do induction on n, for this I considered the following

basic step

Let $n=1$, then $x_{1+2}=x_3=\frac{1}{2}(x_1+x_2)=\frac{1}{2}(2+1)=\frac{1}{2}(3)=\frac{3}{2}$, and clearly the equality is verified.

Now, for the inductive step, we can consider as hypothesis of induction that $1 \leq x_n \leq 2$ with $x_{n+2}=\frac{1}{2}(x_n+x_{n+1})$, $x_1=1$ and $x_2=2$

Thus, let $x_{(n+1)+2}=x_{n+3}=\frac{1}{2}(x_{n+1}+x_{n+2})=\frac{1}{2}(x_{n+1}+\frac{1}{2}( x_n + x_{n+1}))$

Now, it is not very clear to me how to accommodate the inequality to conclude the demonstration, any suggestions?

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2 Answers 2

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HINT

Use strong induction to establish that $x_n$ and $x_{n+1}$ are in $[1,2]$ and then just bound $$ x_{n+2} = \frac{x_n+x_{n+1}}{2}, $$ which is their average...

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It is the average of the two preceding numbers, and avg is between them. So after 2 more iterations you get numbers between 1 and 2, and so avg will be between them and between 1 and 2. And so on and so forth.

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