If you will look at each row down to row 15, you will see that this is true. ((n-1)!)/(1!(n-2)!) You might want to be familiar with this to understand the fibonacci sequence-pascal's triangle relationship. If we look closely at the Pascal triangle and represent it in a combination of numbers, it will look like this. Each row represent the numbers in the powers of 11 (carrying over the digit if it is not a single number). There are various methods to print a pascal’s triangle. First, the outputs integers end with .0 always like in . Using the … Pascal’s triangle can be created as follows: In the top row, there is an array of 1. Look at the 4th line. ; Inside the outer loop run another loop to print terms of a row. Write an expression to represent the sum of the numbers in the nth row of Pascal’s triangle. This works till you get to the 6th line. \({n \choose k}= {n-1 \choose k-1}+ {n-1 \choose k}\) The nth row of Pascal's triangle is: ((n-1),(0)) ((n-1),(1)) ((n-1),(2))... ((n-1), (n-1)) That is: ((n-1)!)/(0!(n-1)!) ; To iterate through rows, run a loop from 0 to num, increment 1 in each iteration.The loop structure should look like for(n=0; n