<s>[INST] is there a formula for this binomial sum? Are there any formulas to calculate the below sum?
$$\sum_{n=1}^{1918}n\binom{2017-n}{99}$$
Or, more generally,
$$\sum_{n=1}^{1918}n\binom{2017-n}{k}$$
 [/INST] A general solution where $p = 2017$ and $k = 99$ in your case:
First rewrite the sum to look nicer:
\begin{align*}
\sum_{n=0}^{p-k}n\binom{p-n}k &= \sum_{n=k}^p(p-n)\binom{n}k = p\sum_{n=k}^p\binom{n}k - \sum_{n=k}^p n\binom{n}k \\
&= p\sum_{n=0}^{p-k}\binom{n+k}k - \sum_{n=0}^{p-k}(n+k)\binom{n+k}k \\
&= (p-k)\sum_{n=0}^{p-k}\binom{n+k}k - \sum_{n=0}^{p-k}n\binom{n+k}k
\end{align*}
Now the nice thing about this is that this looks like a well known generating function! We have that
$$
\frac1{(1-x)^{k+1}} = \sum_{n\geq0}\binom{n+k}kx^n. 
$$
To find the generating function for $n\binom{n+k}k$, we can differentiate both sides and multiply by $x$.
$$
\frac{(k+1)x}{(1-x)^{k+2}} = \sum_{n\geq0}n\binom{n+k}kx^{n-1}\cdot x = \sum_{n\geq0}n\binom{n+k}k x^n
$$
But this is only the generating function for each of the terms, not the sum up to something. The trick for getting the generating function for the first $n$ terms of a sequence from the original generating function $A(x)$ is shifting the terms and adding them up:
$$
A(x) + xA(x) + x^2A(x) + \dots = A(x)\left(1+x+x^2+\dots\right) = \frac{A(x)}{1-x}. 
$$
So, the answer you want is the $(p-k)$th coefficient of the generating function
$$
\frac{p-k}{(1-x)^{k+2}} - \frac{(k+1)x}{(1-x)^{k+3}}. 
$$
The $n$th coefficient of the first term is $(p-k)\binom{n+k+1}{k+1}$ from before, and the $n$th coefficient of the second term is the $n-1$th coefficient of $\frac{(k+1)}{(1-x)^{k+3}}$, which is $(k+1)\binom{n+k+1}{k+2}$. So, we conclude that the answer is
\begin{align*}
(p-k)\binom{p+1}{k+1}-(k+1)\binom{p+1}{k+2} &= (p-k)\frac{k+2}{(p+1)-(k+1)}\binom{p+1}{k+2}-(k+1)\binom{p+1}{k+2} \\
&= ((k+2)-(k+1))\binom{p+1}{k+2} \\
&= \boxed{\binom{p+1}{k+2}.}
\end{align*}
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