Given the Recurrences $T(n)=T(n/2)+2^n$ and $T(n)=T(n/2+\sqrt n)+\sqrt{6044}$
Remark : $T(n)=1$ for $n\le 3$
I'm trying to find their upper bound & lower bound , which is probably $O(2^n)$ for the first one.
I've tried to guess the solution for the first ($T(n)=T(n/2)+2^n$) but it doesn't work , afterwards I've tried the place $m = 2^n$ hence $n=\log(m)$ and use the new equation but still it won't work .
For the second ($T(n)=T(n/2+\sqrt n)+\sqrt{6044}$) I'm trying to guess that $T(n)=O(n)$ , hence $T(n)≤c\cdot n$ , but it still doesn't work.
Any hints and/or directions would be much appreciated .
Regards
EDIT:
About the second one :
$T(n)≤c(n/2+√n)+√6044=cn/2+c√n+√6044=(cn-cn/2)+c√n+√6044= cn-cn/2+c√n+√6044=cn-(cn/2-c√n-√6044) ≤^? cn$
Which is true only if $(cn/2-c√n-√6044)>0$ . What do you think , folks ?