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 Alternating Pascal (Posted on 2011-12-08)
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]

 See The Solution Submitted by Jer No Rating

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 Also true for pascal-like tables(spoiler) | Comment 1 of 2
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.

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
 Posted by Steve Herman on 2011-12-08 13:19:26

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