pascal's triangle question?

Question

what is the formula to claculate the sum of the numbers in the nth row of pascals triangle? how did u get it?

Answer 1

1 (assume 1st row)
11 (2nd row)
121
1331
14641
... and so on

you would know that

(x+1)^n = nC0 x^n + nC1 x^n-1 + nC2 x^n-2 + ..... + nCn 1

the pascal numbers are nothing but nC0, nC1,....nCn for any given row of n+1 (think about it)

so for (n+1)th row, putting x=1 we get

2^n = nC0+nC1+....+nCn

so the sum of numbers in nth row is = 2^(n-1)

Answer 2

The sum of the numbers in the nth row of Pascal's Triangle will always be 2^n. By the way, the very top of the pyramid is row zero.

One way of understanding why is because position in the triangle contains a number that indicates the number of paths that you could have taken to arrive at that point. Since each position in the triangle can go down left or down right, you have 2 choices at each position. So, from the very top (row zero) to the next row, you have two places you could move to. When you are on the next row and you ask yourself, how many paths could I have taken to get where I am, the answer is 1 because there is only 1 position above you.

So, there are 2 positions in row 1 and each of them have 1 way to get there. Row 1 sums to 2. Now, because the triangle is formed by adding the elements from the row above to get the next row, and, because each element in the row above will lead to 2 elements below, the numbers on the row above will get added in twice. This means that the sum of each row will be double the row above it.

Row 0, 2^0 =1
Row 1, 2^1 =2
Row 2, 2^2 =4
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Row n, 2^n