The recurrence equation I have is: $$ T_n = c_1 + T_{n-1} + 2\sqrt{c_2 + c_1 T_{n-1}} $$ $$ T_0 = a $$ $c_1,c_2,a$ are positive real numbers
I need to somehow convert this into a linear homogeneous recurrence or find a closed form if possible.
I tried backward substitution and after a few steps I gave up as it got too complex. I also looked up a few such questions (involving square root) here and tried all the tricks that I came across, taking $S_n = \sqrt{c_2 + c_1 T_n}$ and what not but that didn't seem to help.
An alternate form I came up with (didn't really help though): $$ T_n = n * c_1 + T_1 + 2\sum_{i=1}^{n-1}{f(T_i}) $$ $$ f(x) = \sqrt{c_2 + c_1 x} $$
And there is another form of the above recurrence which does not have any square roots but I found it impossible to make $T_n$ the subject of formula. I will post that too if needed.
I'd be grateful if someone helped me with this.
We start from the recurrence
$$T_n = c_1 + T_{n-1} + 2\sqrt{c_2 + c_1T_{n-1}}.$$
Multiplying that with $c_1$ - that is legitimate since $c_1 > 0$ - and adding $c_2$ yields
\begin{align} c_2 + c_1 T_n &= c_2 + c_1^2 + c_1 T_{n-1} + 2c_1 \sqrt{c_2 + c_1 T_{n-1}}\\ &= c_1^2 + 2c_1\sqrt{c_2 + c_1 T_{n-1}} + (c_2 + c_1 T_{n-1})\\ &= (c_1 + \sqrt{c_2 + c_1 T_{n-1}})^2. \end{align}
Setting $S_n = \sqrt{c_2 + c_1 T_{n}}$, that recurrence is
$$S_n^2 = (c_1 + S_{n-1})^2,$$
which, after taking square roots - no problem because everything is positive - becomes
$$S_n = c_1 + S_{n-1}.$$
With the initial value $S_0 = \sqrt{c_2 + c_1 a}$, we thus have
$$S_n = n\cdot c_1 + \sqrt{c_2 + c_1 a}.$$