giving $f(n)= \sum_{t=1}^n {2(t-1) \choose n-1}$ we need to find a function $g(n)$ such that $f(n)=\theta (g(n))$
I tried with Stirling's approximation but i succeed to find only upper bound which is $\sqrt{n}4^n$.
giving $f(n)= \sum_{t=1}^n {2(t-1) \choose n-1}$ we need to find a function $g(n)$ such that $f(n)=\theta (g(n))$
I tried with Stirling's approximation but i succeed to find only upper bound which is $\sqrt{n}4^n$.
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