Direct proof using telescoping series for $a_n = 1 + 2^{2-n}$

Question

I am practicing for my math exam, I encountered the following problem:

$a_1=3, 2a_n=1+a_{n-1}$ for $n \geq 2$. Give a proof using mathematical induction and direct proof involving telescoping series that $a_n = 1 + 2^{2-n}$.

I was able to do the problem using mathematical induction. However, I am uncertain about the direct proof.

I assume I have to write $a_n$ as an infinite expression. Something like $\frac{1 + \frac{1 + a_{n-2}}{2}}{2}$. I am not sure how to take it from there.

Answer 1

For $n=1$ it's true.

Now, by the assumption of the induction we obtain: $$a_{n+1}=\frac{1+a_n}{2}=\frac{1+1+2^{2-n}}{2}=1+2^{1-n}=1+2^{2-(n+1)}.$$ Also, $$2(a_n-1)=a_{n-1}-1,$$ which gives $$a_n-1=(3-1)\left(\frac{1}{2}\right)^{n-1}.$$

Answer 2

$$a_1=3$$

$$2a_2=1+a_1$$

$$2^2a_3=2+2a_2$$

$$2^3a_4=2^2+2^2a_3$$

$$...$$

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

Summing up and canceling terms:

$$2^{n-1}a_n=3+\sum_{i=0}^{n-2}2^i=2+2^{n-1}$$

and divide by $2^{n-1}$:

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