Q: What is the sum of all the numbers in row 10 of a pascal triangle?

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The Fifth row of Pascal's triangle has 1,4,6,4,1. The sum is 16.

It is 2n, if that is what you were trying to ask.

The numbers are 100Cn = 100!/[n!*(100-n)!] for n = 0, 1, ... , 100

Each element of a row of pascal's triangle is the sum of the two elements above it. Therefore when you some the elements of a row, each of the elements of the row above are being summed twice. Thus the sum of each row of pascal's triangle is twice the sum of the previous row.

Related questions

The sum of the numbers on the fifteenth row of Pascal's triangle is 215 = 32768.

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.)

64

4

All counting numbers. In fact the second number in each row forms the sequence 1,2,3,4,...

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.

If you consider row 0 as the row consisting of the single 1, then row 100 has 6 odd numbers.

The Fifth row of Pascal's triangle has 1,4,6,4,1. The sum is 16.

1 6 15 20 15 6 1

It is 2n, if that is what you were trying to ask.

When evaluating row n+1 of Pascal's triangle, each number from row n is used twice: each number from row ncontributes to the two numbers diagonally below it, to its left and right. From this it is easily seen that the sum total of row n+1 is twice that of row n. The first row of Pascal's triangle, containing only the single '1', is considered to be row zero. Its total, 1, is given by 20. From the above observations, we can conclude that the total of row n is given by 2n. For the eleventh row: 211 = 2048.

The numbers are 100Cn = 100!/[n!*(100-n)!] for n = 0, 1, ... , 100