Solve $$T(n)=T\left(\frac{n}{2}\right)+2^{n}.$$
My approach
\begin{align*} T(n)&=T\left(\frac{n}{2}\right)+2^{n}\\ &=T\left(\frac{n}{4}\right)+2^{\frac{n}{2}}+2^{n}\\ &=T\left(\frac{n}{8}\right)+2^{\frac{n}{2^{2}}}+2^{\frac{n}{2}}+2^{n}, \end{align*}
So $$T(n)=2^{n}+2^{\frac{n}{2}}+2^{\frac{n}{2^{2}}}+\cdots+1. $$
Assuming base case as $T(1)=1$, then
$$T\left(\frac{n}{2^{k}}\right)=1\Longrightarrow \frac{n}{2^{k}}=1 \Longrightarrow k=\log_2{n}.$$
\begin{align*} T(n)&=2^{n}+2^{\frac{n}{2}}+2^{\frac{n}{2^{2}}}+\cdots+1\\ &=2^{n} \times \frac{(2^{\frac{n}{2}})^{k+1}-1}{2^{\frac{n}{2}}-1}. \end{align*}
I am unable to move forward. I know I have to apply geometric progression. Please help me out.