A problem I am doing eventually requires me to try and find a nice, closed form for sums that look like $$ C(3,0) (7-7x)^0 (7-7x+6b)^2 + C(4,1) (7-7x)^1 (7-7x+6b)^1 + C(5,2) (7-7x)^2 (7-7x+6b)^0. $$
(I am doing it in general, but this small example should illustrate the issue).
If not for the changing binomial coefficients, this looks like the standard expansion of $(x+y)^n$.
I have used the identity $$C(m,n) = \sum_{k=0}^n C(m-n-1+k,k) $$ elsewhere in the problem, and it might still be relevant.
I should mention that the problem actually presents as summing two quasi-binomial expansions. Although a closed form for the above would be good, perhaps the problem is only soluble in its entirety. To be exact, the problem (for some values) looks like:
$$ \left(\frac{b}{7x+b} \right) \left(\frac{6b}{7(1-x)+6b} \right)^4 \sum _{k=0}^2 \left[ C(4+k-1,k) \left(\frac{7(1-x)}{7(1-x)+6b} \right)^k \right] + \left(\frac{7x}{7x+b} \right) \left(\frac{6b}{7(1-x)+6b} \right)^5 \sum _{k=0}^1 \left[ C(5+k-1,k) \left(\frac{7(1-x)}{7(1-x)+6b} \right)^k \right] $$ and what I am interested in is showing that this function is (locally) minimised at $x=1/7$.
As I said, I am doing this in general so solutions particular to this sum with just three terms are not quite what I need guidance with. Thanks for your help!
(PS. Mathematica confirms that the function is minimised at $x=1/7$, and the derivative comes to have quite a nice factored form. On the other hand, put the original function (ie those two sums) into mathematica and it spits out something not very helpful. I think the issue is that mathematica expands only into $x's$ and $b's$ but the useful thing is probably to keep much of it in terms of $x^m(1-x)^n$.)