We are given the expression:
$$\binom{98}{30}+2\binom{97}{30}+3\binom{96}{30}+…+68\binom{31}{30}+69\binom{30}{30}=\binom{100}{q}$$
where $q$ is a positive integer $\leq 50$ and we need to find the value of $q$.
I tried evaluating the expression by writing it as a general term and trying to use summation but this presents a problem as it is not like other binomial based sums that I've calculated which were within bounds like $r=0$ to $n$.
The general term that I came up with was $$\sum_{i=1}^{69}i\binom{99-i}{30}$$
But this didn't strike me as anything impressive as I still couldn't figure out how to proceed because the series doesn't seem to simplify out to a form where I can use telescoping or other techniques.
Any help would be appreciated.
Best Answer
Let $[x^p]~f(x)$ mean the co-efficient of $x^{p}$ in $f(x)$.
$$S=\sum_{i=0}^{n} i {m-i \choose p} =[x^p] \sum_{i=0}^{n} i (1+x)^{m-i}$$ The above series is AGP, using its summation, we get $$S=[x^p]~(1+x)^{m-n}\left(\frac{-1-x+nx+(1+x)^n+x(1+x)^n}{x^2}\right)$$ $$\implies [x^{p+2}]~(1+x)^{m-n}(-1-x+nx+(1+x)^n+x(1+x)^n)$$ $$\implies S=-{m-n \choose p+2}+(n-1) {m-n \choose p+1}+{m \choose p+2}+{m \choose p+1}$$ $$S=-{m-n+1 \choose p+2}+n{m-n \choose p+1}+{m+1 \choose p+2}$$ For the given problem $m=99, n=69, p=30$, then the first twp coefficirnt disappear and we get $$S={100 \choose 32}.$$ Finally, $q=32$.