Row 22 of Pascal's Triangle consists of the binomial coefficients (\binom{22}{0}), (\binom{22}{1}), (\binom{22}{2}), ..., (\binom{22}{22}). The values in this row are: 1, 22, 231, 1540, 7315, 26334, 7315, 1540, 231, 22, and 1. Thus, row 22 is: 1, 22, 231, 1540, 7315, 26334, 497420, 497420, 26334, 7315, 1540, 231, 22, 1.
To convert 760 torr to pascals, you can use the fact that 1 torr is equal to 133.322 pascals. Therefore, you multiply 760 torr by 133.322 pascals/torr: 760 torr × 133.322 pascals/torr = 101325 pascals. Thus, 760 torr is equivalent to 101325 pascals.
Pascal's triangle is created by starting with a single "1" at the top, known as the apex. Each subsequent row is formed by adding the two numbers directly above it from the previous row. If there is no number above (for edge cases), a "0" is assumed. This process continues indefinitely, with each row corresponding to the coefficients of the binomial expansion.
A hectopascal is 100 pascals
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No
1,4,6,4,1
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.
The sum of the 17th row of Pascal's Triangle can be calculated using the formula 2^n, where n is the row number minus one. In this case, the 17th row corresponds to n=16. Therefore, the sum of the 17th row is 2^16, which equals 65,536.
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.
To find (10C3) using Pascal's Triangle, locate the row corresponding to (n=10). The entries in this row represent the binomial coefficients for (n=10). The third entry (starting from (0)) in this row corresponds to (10C3), which is (120). Thus, (10C3 = 120).