Algebra II, Pascal's Triangle.?

Question

I'm having a little trouble with this problem about pascal's triangle.

I'm supposed find the connection between the number of rows in the triangle, and sum of all numbers in the triangle.

For example: If there are 2 rows in in the triangle, the sum of all the numbers in the triangle would be 3.

So... i have to find the the sum of all the numbers in the triangle if I had 50 rows in the triangle.
Also the same thing with "n"

If anybody has any ideas, please post =]]

Answer 1

The horizontal rows of a Pascal's triangle add to powers of 2. For example, the first row adds up to 1, that is 2^0, the second adds up to 2, that is 2^1, the third to 4 (2^2), the fourth to 8 (2^3) and so on.

So, row 'n' adds up to 2^(n-1).

Which means that the sum of all the numbers in the triangle if there are 'n' rows in the triangle is

2^0 + 2^1 + 2^2 + 2^3 + ... + 2^(n-1)

Which is given by (2^n - 1).

When n=50, the sum is 2^50 - 1.

Answer 2

speaking in binary (base 2) the answers are as follows...
1
11
111
1111
11111
111111
remember that in binary the first digit from the right is 2^0 and the next moving left is 2^1, then 2^2 and so on. if there is a 1 in the value place then it is worth the total of the place value, if not then it is worth 0. thus,

111111 would be the same as
32+16+8+4+2+1 thus 63 or the value of the next place value minus '1' thus the 50th row would have the value of 2^(n-1) or you could replace 'n' with '50'

Answer 3

hmm ive been thinking about it and ive noticed that the difference of the sum of the first row and 2nd is 2.
then the 2nd and 3rd is 4
then the 3rd and 4th is 8
and the 4th and 5th is 16.

basically with each succeeding term, the difference doubles but i cant seem to figure it out.. ill email you when i crack the case ;)