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This is the recursive formula: $$\begin{cases} a_1=1/2,\\ a_n= \sqrt {\frac {a_{n-1}+1}{2}} \end{cases}$$

I have calculated the first 4 terms to be $\frac12, \frac {\sqrt3}2, \frac {\sqrt{\sqrt3+2}}2, \frac {\sqrt{\sqrt{\sqrt3+2}+2}}2$

How can I find the explicit formula?

I have found that for $b_n=\frac {a_{n-1}+1}{2}$, the explicit equation would be: $$b_n=-\left(\frac12\right)^n+1$$ However, I am struggling to find the explicit equation when the square root is included.

I was thinking since $\sqrt b_n=a_n$, then maybe I could equate $a_n$ with $\sqrt {-\left(\frac12\right)^n+1}$, but unfortunately the graph does not line up with the points.

This is the recursive formula: $$\begin{cases} a_1=1/2,\\ a_n= \sqrt {\frac {a_{n-1}+1}{2}} \end{cases}$$

I have calculated the first 4 terms to be $\frac12, \frac {\sqrt3}2, \frac {\sqrt{\sqrt3+2}}2, \frac {\sqrt{\sqrt{\sqrt3+2}+2}}2$

How can I find the explicit formula?

Looking at the first 4 terms, I notice that the denominator remains at 2, while the numerator appears to also be recursive. I will call this numerator $b_n$ $$\begin{cases} b_1=1,\\ b_n= \sqrt {b_{n-1}+2} \end{cases}$$

I am thinking that if I find the explicit formula for $b_n$, perhaps I could use the formula $\frac {b_n}2=a_n$ and solve.

I also figure out that in the recursive formula where $c_1=x$, and $c_{n}=\sqrt{c_{n-1}}$, the explicit formula would be: $$c_n=x^\frac1{2^{(n-1)}}$$

I am currently struggling on how to solve for the case $b_n=\sqrt{b_{n-1}+2}$

This is the recursive formula: $$\begin{cases} a_1=1/2,\\ a_n= \sqrt {\frac {a_{n-1}+1}{2}} \end{cases}$$ How can I find the explicit formula?

I have found that for $b_n=\frac {a_{n-1}+1}{2}$, the explicit equation would be: $$b_n=-\left(\frac12\right)^n+1$$ However, I am struggling to find the explicit equation when the square root is included.

I was thinking since $\sqrt b_n=a_n$, then maybe I could equate $a_n$ with $\sqrt {-\left(\frac12\right)^n+1}$, but unfortunately the graph does not line up with the points.

This is the recursive formula: $$\begin{cases} a_1=1/2,\\ a_n= \sqrt {\frac {a_{n-1}+1}{2}} \end{cases}$$

I have calculated the first 4 terms to be $\frac12, \frac {\sqrt3}2, \frac {\sqrt{\sqrt3+2}}2, \frac {\sqrt{\sqrt{\sqrt3+2}+2}}2$

How can I find the explicit formula?

I have found that for $b_n=\frac {a_{n-1}+1}{2}$, the explicit equation would be: $$b_n=-\left(\frac12\right)^n+1$$ However, I am struggling to find the explicit equation when the square root is included.

I was thinking since $\sqrt b_n=a_n$, then maybe I could equate $a_n$ with $\sqrt {-\left(\frac12\right)^n+1}$, but unfortunately the graph does not line up with the points.

This is the recursive formula: $$\begin{cases} a_1=1/2,\\ a_n= \sqrt {\frac {a_{n-1}+1}{2}} \end{cases}$$ How can I find the explicit formula?

I have found that for $b_n=\frac {a_{n-1}+1}{2}$, the explicit equation would be: $$b_n=-\left(\frac12\right)^n+1$$ However, I am struggling to find the explicit equation when the square root is included.

This is the recursive formula: $$\begin{cases} a_1=1/2,\\ a_n= \sqrt {\frac {a_{n-1}+1}{2}} \end{cases}$$ How can I find the explicit formula?

I have found that for $b_n=\frac {a_{n-1}+1}{2}$, the explicit equation would be: $$b_n=-\left(\frac12\right)^n+1$$ However, I am struggling to find the explicit equation when the square root is included.

I was thinking since $\sqrt b_n=a_n$, then maybe I could equate $a_n$ with $\sqrt {-\left(\frac12\right)^n+1}$, but unfortunately the graph does not line up with the points.