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For $n\in\mathbb{N}$, the binomial theorem gives $$\sum_{j=0}^n\binom{n}{j}x^j=(1+x)^n$$ I am interested in the case when the series is incomplete, and specifically for $x=2$, namely: $$r_{n,k}=\sum_{j=0}^{k}\binom{n}{j}2^j$$ In this question, I established the formula $$r_{n,k}=3^n/2^k-2 {n\choose k+1} {}_{2}F_1(1,k-n+1;k+2;-2),$$ where ${}_2F_1$ is the usual hypergeometric function. Similar questions have asked about different cases (for instance, this question is the case $x=-1$, which gives a nice formula in terms of the limit of summation; this MO question asks about the asymptotics for $k\le n/2$), but again I'm focused in the full range of the upper sum in the case $x=2$. Perhaps the answer lies in manipulating the hypergeometric function into something more tangible.

kc9jud
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Integrand
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  • Can't you factor $2^k$ out of the sum? – kc9jud Sep 11 '24 at 17:44
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    Presumably, you meant $2^j,$ not $2^k$ in your definition of $r_{n,k}?$ – Thomas Andrews Sep 11 '24 at 17:45
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    Do you really mean the case $k=2,$ rather than the case $x=2?$ – Thomas Andrews Sep 11 '24 at 17:48
  • Apologies; I changed variables several times and a few times typos persisted – Integrand Sep 11 '24 at 17:53
  • Dividing both sides by $2^k,$ you get that $r_{k,n}/2^k$ is is the coefficient of $x^k$ of $f_n(x)=\frac{(1+x)^n}{1-x/2}.$ Not sure that helps. You can definitely get a recursion of some sort as $n$ increasing by writing $f_n(x)=p_n(x)+\frac{b_n}{1-x/2}.$ Then $$f_{n+1}(x)=(x+1)p_n(x)+p_1(x)+\frac{b_1b_n}{1-x/2}$$ or something like that. – Thomas Andrews Sep 11 '24 at 17:58
  • Asking Mathematica to attack the series in your second display indicates the left-hand side of your third display should be $r_{n,k}/2^k$ or that the right-hand side should start "$3^n - 2^{k+1}\dots$". – Eric Towers Sep 11 '24 at 18:15
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    Letting $X$ be a Binomial $(n,p)$ with $p=2/3$, we get $$r_{n,k} =3^{n} P(X \le k) $$ Plugging the CLT approximation, this should give a decent asymptotic for $k \sim n \frac 23$ – leonbloy Sep 11 '24 at 18:29

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