I encountered something similar to this on a quiz and really had no clue how to proceed. Our in-class problem did not have the A+B-C thing but was pretty similar to this
$$ \sum_{n=1}^{\infty}{\frac{\prod_{r=0}^{n-1}{(\pi-r})}{2^nn!}}=\left(\frac{A}{B}\right)^{\pi}-C$$ If the equation above holds for coprime integers $A,B,\text{ and }C,$ enter the value of $A+B-C$
https://brilliant.org/problems/crazy-sum-2/
I knew the first step was to use a binomial series, but did not know how to proceed from there. Any suggestions as to how to proceed/solutions?
The whole point of the problem is to recognize the binomial series. $$(1+x)^{\alpha}=\sum_{n=0}^{\infty}{\binom{\alpha}{n}x^n}, $$ where $\alpha$ is a complex number and the (generalized) binomial coefficient $$ \binom{\alpha}{n}=\frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)}{n!} $$ Rewrite the original equations as $$ \sum_{n=1}^{\infty}{\binom{\pi}{n}2^{-n}}=\left(\frac{A}{B}\right)^{\pi}-C $$ Compare this to the binomial formula. We see that $x=1/2$ and that we are missing the $n=0$ term, which is $1,$ from the sum; that is the sum is equal to $$(1+1/2)^{\pi}-1\implies A=3, B=2, C= 1, A+B-C=4$$
In the future, please take the trouble to typeset your questions in MathJax.