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 \(2^{k}\).