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Sums of Rows in Pascal's Triangle


Firstly, I have written out the first few rows of Pascal's triangle and calculated the sums of all the numbers in each row so that we can see if there is a pattern:

1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

1+1=2
1+2+1=4
1+3+3+1=8
1+4+6+4+1=16
1+5+10+10+5+1=32

You should begin to see a pattern emerging: The sums of the rows are the doubling numbers 2,4,8,16,32, where each number is twice the previous one. Mathematically, we could write the sum of row n is 2^n (this means 2x2x2... n times. For example, 2^5 = 2x2x2x2x2, and 2^3 = 2x2x2.

It's all very well spotting this intriguing pattern, but this alone is not entirely satisfactory for a mathematician. How do we know that this pattern always works? Checking it against more rows will not help because however many rows we check, we cannot be sure it will work for the next one.

We need a mathematical proof. It is this act of showing beyond any doubt something to be true or not true by a series of purely logical steps that sets mathematics apart from the other sciences. When we prove something mathematically, we can be absolutely certain it is always true, unlike a practical scientist who will carry out experiments (like our tests in the first diagram), and thus be only at most fairly certain of their results.

We are going to prove (informally) this by a method called induction. This may sound scary, but in this case, its simple. Take any row on Pascal's triangle, say the 1, 4, 6, 4, 1 row. Now think about the row after it. We can write down the next row as an uncalculated sum, so instead of 1,5,10,10,5,1, we write 0+1, 1+4, 4+6, 6+4, 4+1, 1+0. You should be able to see that each number from the 1, 4, 6, 4, 1 row has been used twice in the calculations for the next row. So we start with 1, 1 on row one, and each time every number is used twice in the following row, and hence the total of the rows of Pascal's triangle always doubles.

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