Bi...

Relevant wiki: Binomial Coefficient

Pascal's rule states that:

$\begin{aligned} {x \choose y} + {x \choose y+1} & = {x+1 \choose y+1} \\ \implies {x \choose y} & = {x+1 \choose y+1} - {x \choose y+1} \\ {x \choose y} + {\color{#3D99F6} {x+1 \choose y}} & = \underbrace{\color{#D61F06}{\color{#3D99F6} {x+1 \choose y}} + {x+1 \choose y+1}}_{\color{#D61F06} \text{By Pascal's rule}} - {x \choose y+1} \\ & = {\color{#D61F06} {x+2 \choose y + 1}} - {x \choose y+1} \\ {x \choose y} + {x+1 \choose y} + {\color{#3D99F6} {x+2 \choose y}} & = {\color{#D61F06} {x+3 \choose y + 1}} - {x \choose y+1} \\ {x \choose y} + {x+1 \choose y} + {x+2 \choose y} + {\color{#3D99F6} {x+3 \choose y}} & = {\color{#D61F06} {x+4 \choose y + 1}} - {x \choose y+1} \\ \cdots \ & = \ \cdots \\ {x \choose y} + {x+1 \choose y} + {x+2 \choose y} + {x+3 \choose y} + \cdots + {\color{#3D99F6} {x+z \choose y}} & = {\color{#D61F06} {x+z+1 \choose y + 1}} - {x \choose y+1} \end{aligned}$

For $x=2017$ and $y=100$ , we have $n = \boxed{\displaystyle {2017 \choose 101}}$ .

The hockey stick identity states that

$\sum_{i = n}^m \binom{i}{n} = \binom{m + 1}{n + 1}.$

This identity can be proven using Pascal's identity, as per Chew-Seong's solution.

By the identity, we have

$\begin{aligned} \binom{2017}{100} + \binom{2018}{100} + \cdots + \binom{3018}{100} &= \sum_{i = 2017}^{3018} \binom{i}{100} \\ &= \sum_{i = 100}^{3018} \binom{i}{100} - \sum_{i = 100}^{2016} \binom{i}{100} \\ &= \binom{3019}{101} - \boxed{\dbinom{2017}{101}}. \end{aligned}$

Note that

$\begin{aligned} \binom {n}{r} + \binom {n}{r-1} &= \dfrac{n!}{r!(n-r)!} + \dfrac{n!}{(r-1)!(n-r+1)!} \\ &= \dfrac{n!}{(r-1)!(n-r)!} \left( \dfrac 1r + \dfrac{1}{n-r+1} \right) \\ &= \dfrac{n!}{(r-1)!(n-r)!} \left( \dfrac{n+1}{r(n+1-r)} \right) \\ &= \dfrac{(n+1)!}{r! (n+1-r)!} \\ &= \binom{n+1}{r} \end{aligned}$

$\therefore \dbinom{n}{r} = \dbinom{n+1}{r} - \dbinom{n}{r-1}$

Following the generalization given in the problem

$\begin{aligned} & \binom{x}{y} + \binom{x+1}{y} + \cdots + \binom{x+z}{y} = \binom{x+z+1}{y+1} - k \\ \implies & k = \left[ \binom{x+z+1}{y+1} - \binom{x+z}{y} \right] - \binom{x+z-1}{y} - \cdots - \binom{x+1}{y} - \binom{x}{y} \\ \implies & k = \left[ \binom{x+z}{y+1} - \binom{x+z-1}{y} \right] - \cdots - \binom{x+1}{y} - \binom{x}{y} \\ & \vdots \\ \implies & k = \binom{x+1}{y+1} - \binom{x}{y} \\ \implies & k = \binom{x}{y+1} \end{aligned}$

Thus for $x=2017 \ , \ y=100$ we have $\boxed{n= \dbinom{2017}{101}}$ .