Answer to Question #185251 in Discrete Mathematics for Evans

Question

Generate up to the seventh and nineth rows of the Pascal triangle.

Answer 1

Pascal triangle starts with two initial rows: the 0th row and the 1st row. They are

$\begin{array}{ccc}&1&\\1& &1\end{array}$

All entries in any row are numbered from 0.

The rule to construct the 2nd row is following. Put 1 as the 0th entry, calculate the 1st entry as the sum of 1+1, where each "1" is taken from previous row, then put 1 as the 2nd entry. Thus, Pascal triangle containing 3 rows is

 

$\begin{array}{ccccc}&&1&&\\&1& &1&\\1&&2&&1\end{array}$

 

Any other row starts with 1 and ends with 1. Any other entry within a row lying "between" and "below" two entries of upper row is the sum of the two entries. Thus, the 3rd row is 1, 1+2, 2+1, 1. Triangle containing 4 rows is

$\begin{array}{cccccccc}&&&1&&&\\&&1& &1&&\\&1&&2&&1&\\1&&3&&3&&1\end{array}$  

 

Consequently the 4th row is 1, 1+3, 3+3, 3+1, 1.

 

Continuing up to ninth row construct Pascal triangle.

$\begin{array}{ccccccccccccccccccc}&&&&&&&&&1&&&&&&&&&\\&&&&&&&&1& &1&&&&&&&&\\&&&&&&&1&&2&&1&&&&&&&\\&&&&&&1&&3&&3&&1&&&&&&\\&&&&&1&&4&&6&&4&&1&&&&&\\&&&&1&&5&&10&&10&&5&&1&&&&\\&&&1&&6&&15&&20&&15&&6&&1&&&\\&&1&&7&&21&&35&&35&&21&&7&&1&&\\&1&&8&&28&&56&&70&&56&&28&&8&&1&\\1&&9&&36&&84&&126&&126&&84&&36&&9&&1\end{array}$