If you are talking about the 6th numerical row (1 5 10 10 5 1, technically 5th row because Pascal's triangle starts with the 0th row), it does not appear to be a multiple of 11, but after regrouping or simplifying, it is. We have 1 5 10 10 5 1 which is equivalent to 105 + 5*104 + 10*103 + 10*102 + 5*101 + 1 = 161051. This is equal to 115.
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1,4,6,4,1
The sum is 24 = 16
If the top row of Pascal's triangle is "1 1", then the nth row of Pascals triangle consists of the coefficients of x in the expansion of (1 + x)n.
1 5 10 10 5 1
To find a specific number in Pascal's triangle, use the formula n!/r!(n-r)! Where n is the row number (starting at 0) and r is the row element (starting at 0) So for the 3rd entry of the 12th row we would put in the following numbers: 11!/2!(11-2)! which equals 55 Do the same for the 6th entry of the 12th row and you get 462.