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Number theory Difficulty 4.7 AIME Find the answer

Pascal has a triangle. In the nnth row, there are n+1n+1 numbers an,0,an,1,an,2,…,an,na_{n, 0}, a_{n, 1}, a_{n, 2}, \ldots, a_{n, n} where an,0=an,n=1a_{n, 0}=a_{n, n}=1. For all 1≤k≤n−1,an,k=an−1,k−an−1,k−11 \leq k \leq n-1, a_{n, k}=a_{n-1, k}-a_{n-1, k-1}. What is the sum of all numbers in the 2018th row?

A number or a short expression. Spacing and $ signs are ignored.

Solution

In general, the sum of the numbers on the nnth row will be ∑k=0nan,k=an,0+∑k=1n−1(an−1,k−an−1,k−1)+an,n=an,0+(an−1,n−1−an−1,0)+an,n=2\sum_{k=0}^{n} a_{n, k}=a_{n, 0}+\sum_{k=1}^{n-1}\left(a_{n-1, k}-a_{n-1, k-1}\right)+a_{n, n}=a_{n, 0}+\left(a_{n-1, n-1}-a_{n-1,0}\right)+a_{n, n}=2

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.