The sum of the n-th row of Pascal's Triangle is always 2 to the power of n.

Yesterday, we discussed the Binomial Theorem, the standard way to expand a binomial, here expressed as (x + y), raised to any positive integer power n.


We also have a simple pattern easily visible in the first few rows of Pascal's Triangle.

1 = 1
2 = 1+1
4 = 1+2+1
8 = 1+3+3+1
16= 1+4+6+4+1
...

From what we can see, the sum of the n-th row is 2 raised to the power of n. In math, we can't just say we see a pattern for a few rows and assume it is always true, we need to prove it is true. Here is the proof.
  

Step 1 is no more difficult than 2 = 1+1.  Pretty easy.

Step 2 is to take the Binomial Theorem and replace the general binomial (x + y) with the specific binomial (1+1). So we get all the entries of row n multiplied by powers of 1 then added up.

Any integer power of 1 is 1, so we can erase those powers because they are superfluous.

Now we just have the sum across the row. Since the equations are of the form

a = b, then b = c, then c = d, we can also say a = d.

This is not the only way to prove this. Another proof of this basic property of Pascal's Triangle tomorrow.