Looking for formula for variation of binomial theorem Is there variation of the binomial theorem as follows?
$$\sum_{i=0}^n {m \choose i} a^{m-i} b^i $$
I am trying to find a formula for it that is a function of $n$ without summation notation. I think you would call this is a solution in closed form. Any help would be great. Thanks.
 A: This may not be what you were hoping for but we certainly can write your sum as a function of $n$ without summation notation using this expression for truncated hypergeometric series. We have
$$
\sum_{i=0}^n\binom{m}{i}a^{m-i}b^i=a^m(-b/a)^n\frac{(-m)_n}{n!}{_2F}_1\left({-n,1 \atop 1-n+m};-\frac{a}{b}\right),
$$
where $(x)_n$ is the Pochhammer symbol. Using transformations for the hypergeometric function may be able to put this in a more compact form.
A: This may or may not be a closed form, but one way you could get rid of the summation notation would be to do BinomialSum=$Σ(a_i)$=$exp(ln(Σ(a_i))$=$exp(Πln(a_i))$= $exp\prod_{i=0}^n ln[\binom{m}{i}a^{m-i}b^i]$=
Then you could expand this out using the definition of the binomial operator:
$exp\prod_{i=0}^n [ln\binom{m}{i}+(m-i)ln (a)+i*ln(b)$]= $exp\prod_{i=0}^n [ln(m!)-ln(i!)-ln((m-i)!)+(m-i)ln (a)+i*ln(b)$]
Then you could use the definition of the factorial: x!=$\prod_{k=1}^{x-1}$(x-k) to then get ln(x!)=
ln$\prod_{k=1}^{x-1}$(x-k)=$\sum_{k=1}^{x-1}(ln(x-k))$
