Maths Olympiad Prep

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, 2018

Combinatorics Difficulty 4.8 AIME Find the answer United States

Problem:

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 1kn11 \leq k \leq n-1, an,k=an1,kan1,k1a_{n, k}=a_{n-1, k}-a_{n-1, k-1}. Let NN be the value of the sum
k=02018a2018,k(2018k). \sum_{k=0}^{2018} \frac{\left|a_{2018, k}\right|}{\binom{2018}{k}} .
Estimate NN.

An estimate of E>0E>0 earns 202NE/70\left\lfloor 20 \cdot 2^{-\left|N-E\right| / 70}\right\rfloor points.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

```python
import math
lists=[[1]]
for i in range(2018):
newlist=[]
for j in range(i):
newlist.append(lists[-1][j+1]-lists[-1][j])
lists.append([1]+newlist+[1])
big=math.factorial(2018)
sum=0
for i in range(2019):
sum+=abs(lists[-1]*)/(big//math.factorial(i)//math.factorial(2018-i))
print(sum)
```

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.