The sum of the numbers in each row of Pascal's triangle is twice the sum of the previous row. Perhaps you can work it out from there. (Basically, you should use powers of 2.)
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 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 is 65,528
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
If you consider row 0 as the row consisting of the single 1, then row 100 has 6 odd numbers.
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.
1 5 10 10 5 1
1,4,6,4,1
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 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 is 65,528
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
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, 9, 36, 84, 126, 126, 84, 36, 9, 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.
To find a specific number in Pascal's triangle, use the formula n!/r!(n-r)! Where n is the row number (starting at 0) and r is the row element (starting at 0) So for the 3rd entry of the 12th row we would put in the following numbers: 11!/2!(11-2)! which equals 55 Do the same for the 6th entry of the 12th row and you get 462.