I get it to be:
1, 25, 301, 2302, 12651, 53130, 177100, 480700, 1081575, 2042975, 3268760, 4457400, 5200300, 5200300, 4457400, 3268760, 2042975, 1081575, 480700, 177100, 53130, 12651, 2302, 301, 25, 1
* * * * *
Some of the figures are incorrect:
1
25
300
2300
12650
53130
177100
480700
1081575
2042975
3268760
4457400
5200300
and then back down
1,4,6,4,1
The sum is 24 = 16
If the top row of Pascal's triangle is "1 1", then the nth row of Pascals triangle consists of the coefficients of x in the expansion of (1 + x)n.
1 5 10 10 5 1
The Fifth row of Pascal's triangle has 1,4,6,4,1. The sum is 16. Formula 2n-1 where n=5 Therefore 2n-1=25-1= 24 = 16.
1,4,6,4,1
the sum is 65,528
depends. If you start Pascals triangle with (1) or (1,1). The fifth row with then either be (1,4,6,4,1) or (1,5,10,10,5,1). The sums of which are respectively 16 and 32.
The sum is 24 = 16
If the top row of Pascal's triangle is "1 1", then the nth row of Pascals triangle consists of the coefficients of x in the expansion of (1 + x)n.
1 5 10 10 5 1
The Fifth row of Pascal's triangle has 1,4,6,4,1. The sum is 16. Formula 2n-1 where n=5 Therefore 2n-1=25-1= 24 = 16.
1, 9, 36, 84, 126, 126, 84, 36, 9, 1
Sum of numbers in a nth row can be determined using the formula 2^n. For the 100th row, the sum of numbers is found to be 2^100=1.2676506x10^30.
Pascal's triangle
The number of odd numbers in the Nth row of Pascal's triangle is equal to 2^n, where n is the number of 1's in the binary form of the N. In this case, 100 in binary is 1100100, so there are 8 odd numbers in the 100th row of Pascal's triangle.
The sum of the numbers in the nth row of Pascal's triangle is equal to 2^n. Therefore, the sum of the numbers in the 100th row of Pascal's triangle would be 2^100. This formula is derived from the properties of Pascal's triangle, where each number is a combination of the two numbers above it.