Équivalent to $\sum_{k=0}^n \binom{2n}{k} $

Question

Hollo Let $u_n = \sum_{k=0}^n \binom{2n}{k} $ then after change indice and pascal relation we have

u_0=1 and. $u_n = 4u_n -\frac{1}{n+1} \binom{2n}{n}$.

We remark. u_n équivalent (2/pi)*4^n

Answer 1

Note that $2u_n=4^n+{2n\choose n}$.

And ${2n\choose n}\sim\dfrac{4^n}{\sqrt{n\pi}}$ (easy to prove with Wallis' integrals or Stirling's formula), hence $u_n\sim 2^{2n-1}$.