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For some integer $z \ge 2$ and large integer $n$ and $ t=\lceil \log n\rceil $, what is an approximate value for the following partial binomial sum? $$ \sum_{i=0}^{n-t} \binom{n}{i}z^i .$$

Another related problem for the same parameters is approximating $$ \sum_{i=n/t}^n \binom{n}{i}z^i .$$

Thanks.

Aryo Z
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    This is close to http://mathoverflow.net/questions/17202/sum-of-the-first-k-binomial-coefficients-for-fixed-n – Suvrit Nov 12 '15 at 05:32
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    In both cases the sum is asymptotic to $(1+z)^n$ with great precision, since you have given almost all of the binomial summation. – Brendan McKay Nov 12 '15 at 05:38
  • @BrendanMcKay true! I did not notice that $t$ is so small and took it to be arbitrary! – Suvrit Nov 12 '15 at 05:42
  • @BrendanMcKay Thanks. Do you any reference or hint for its proof? – Aryo Z Nov 12 '15 at 06:13
  • You can just be crude and say ${n \choose n-m} \leq n^m$ in this range. Thus the sum of the missing terms is at most $t n^t z^n$, which is exponentially smaller than $(1+z)^n$. – alpoge Nov 12 '15 at 07:18
  • The largest term is at $k=nx/(1+x)$. If $f(n)/\sqrt{n}\to\infty$, the terms with $|k -nx/(1+x)| \le f(n)$ give most of the sum. So any interval of summation that includes those important values of $k$ will be asymptotic to $(1+x)^n$. – Brendan McKay Nov 12 '15 at 08:18

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