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If $\langle f_{n}\rangle$ be a sequence of positive numbers such that $$f_{n}=\frac{f_{n-1}+f_{n-2}}{2}$$ $\forall n\gt2$ ,then show that $\lt f_{n}\gt$ converges to $$\frac{f_1+2f_2}{3}$$

Replacing $n$ by $3,4,5,....,n-1$,we get $$f_3=(f_2+f_1)/2$$ $$f_4=(f_3+f_2)/2$$ $$f_5=(f_4+f_3)/2$$ $$\dots\dots\dots\dots\dots$$ $$f_{n-1}=(f_{n-2}+f_{n-3})/2$$ $$f_{n}=(f_{n-1}+f_{n-2})/2$$

In the solution, its given as adding the corresponding sides of $f_n$ and $f_{n-1}$,the following expression is obtained

$$f_n+\frac12f_{n-1}=\frac12(f_1+2f_2)$$

I couldn't understand how this expression came.

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May be be easier to follow if you first rewrite the given recurrence as:

$$f_{n} \color{red}{+\frac{f_{n-1}}{2}}=\frac{f_{n-1}+f_{n-2}}{2} \color{red}{+\frac{f_{n-1}}{2}} = f_{n-1}+\frac{f_{n-2}}{2}$$

It "telescopes" nicely from there on:

$$f_{n} +\frac{f_{n-1}}{2} = f_{n-1}+\frac{f_{n-2}}{2} = f_{n-2}+\frac{f_{n-3}}{2} = \cdots = f_2 + \frac{f_1}{2}$$

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On the right side every number except $f_1, f_2, f_{n-1}$ appear twice and when you multiply by $\frac 12$, each of those number has a coefficient 1 in front of it. Similarly on the left side all numbers appear once, so after massive cancelations you get the wanted identity.

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In this case you can also explicitely solve

$$f_{n}=\frac{f_{n-1}+f_{n-2}}{2}\implies2x^2-x-1=0\implies x_1=1 \quad x_2=-\frac12\implies \\f_n=a+b\left(-\frac12\right)^n\to a$$

and from the initial conditions

$$\begin{cases} f_1=a+b\\ f_2=a-\frac12b \end{cases}\implies 3a=f_1+2f_2\implies a=\frac{f_1+2f_2}{3}$$

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