Graphical representation of the coefficients in Newton’s binomial theorem in which there is a pattern that can be used to calculate any value in a given row.
This is a partial illustration of Pascal’s triangle :
| If n = 0 | 1 | ||||||||||||||||
| If n = 1 | 1 | 1 | |||||||||||||||
| If n = 2 | 1 | 2 | 1 | ||||||||||||||
| If n = 3 | 1 | 3 | 3 | 1 | |||||||||||||
| If n = 4 | 1 | 4 | 6 | 4 | 1 | ||||||||||||
| If n = 5 | 1 | 5 | 10 | 10 | 5 | 1 | |||||||||||
| If n = 6 | 1 | 6 | 15 | 20 | 15 | 6 | 1 | ||||||||||
| If n = 7 | 1 | 7 | 21 | 35 | 35 | 21 | 7 | 1 | |||||||||
| If n = 8 | 1 | 8 | 28 | 54 | 70 | 54 | 28 | 8 | 1 | ||||||||
| … |
Each number is the sum of the two numbers located above it in the triangle.
The sum of the elements in the row corresponding to n = k is \(2^{k}\).