Prove that for any row of Pascal's Triangle if you alternately subtract and add terms you always get 0.
Note: this is obvious for odd rows [1-3+3-1=0]
but not so obvious for even rows [1-4+6-4+1=0]
Consider row n and row (n+1). Every term in row (n+1) is formed by adding the 1 or 2 terms in row n which are above it and to the immediate left and right. Call the 1st and 3rd and 5th term in any row the odd terms. Every number in row n contributes (by addition) to two terms of row (n+1), an even term and an odd term. So, if we subtract the even terms from the odd ones, the contribution in row (n+1) from every row n term nets out to 0. Therefore, if we subtract the even terms from the odd ones in any row, they must net out to 0.
This is true even if we start with a pascal-like triangle.
For instance, start with a random non-pascal row for which the property is not true, like
1 2
Subsequent rows, generated pascal-like, are
1 3 2
1 4 5 2
1 5 9 7 2
And the property holds for all rows generated from the random starter row.
Edited on December 9, 2011, 10:31 am
Edited on December 9, 2011, 10:39 am
Also true for pascal-like tables(spoiler)
