The rth term of the 25th row is 25!/[r!(25-r)!] where r = 0,1,2,...,25 and k! denotes 1*2*3*...*k and 0! = 1
So
1
25
300
2,300
12,650
53,130
177,100
480,700
1,081,575
2,042,975
3,268,760
4,457,400
5,200,300
and then the same numbers in reverse order, all the way back to 1.
pascals triangle is used to solve math problems that have chance of 2 different outcomes, such as flipping a coin
The terms in row 29 are: 29Cr = 29!/[r!*(29-r)!] for r = 0, 1, 2, ... 29 where r! denotes 1*2*3*...*r and 0! = 1
in the 11th century...
its useful if you work as a architect
The sum of the numbers on the fifteenth row of Pascal's triangle is 215 = 32768.
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 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.
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
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.
Each number in Pascal's triangle is used twice when calculating the row below. Consequently the row total doubles with each successive row. If the row containing a single '1' is row zero, then T = 2r where T is the sum of the numbers in row r. So for r=100 T = 2100 = 1267650600228229401496703205376