Question

How do the values in Pascal’s triangle connect to the coefficients?

Answer 1

Explanation:

Each row in Pascal's triangle is a listing of the values of nCk = n!/(k!(n-k)!) for some fixed n and k in the range 0 to n. nCk is the number of combinations of n things taken k at a time.

If you consider what happens when you multiply out the product (a +b)^n, you can see where the coefficients nCk come from. For example, consider the cube ...

  (a +b)^3 = (a +b)(a +b)(a +b)

The highest-degree "a" term will be a^3, the result of multiplying together the first terms of each of the binomials.

The term a^b will have a coefficient that reflects the sum of all the ways you can get a^b by multiplying different combinations of the terms. Here they are ...

• (a +_)(a +_)(_ +b) = a·a·b = a^2b
• (a +_)(_ +b)(a +_) = a·b·a = a^2b
• (_ +b)(a +_)(a +_) = b·a·a = a^2b

Adding these three products together gives 3a^2b, the second term of the expansion.

For this cubic, the third term of the expansion is the sum of the ways you can get ab^2. It is essentially what is shown above, but with "a" and "b" swapped. Hence, there are 3 combinations, and the total is 3ab^2.

Of course, there is only one way to get b^3.

So the expansion of the cube (a+b)^3 is ...

  (a +b)^3 = a^3 + 3a^2b +3ab^2 +b^3 . . . . . with coefficients 1, 3, 3, 1 matching the 4th row of Pascal's triangle.

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In short, the values in Pascal's triangle are the values of the number of combinations of n things taken k at a time. The coefficients of a binomial expansion are also the number of combinations of n things taken k at a time. Each term of the expansion of (a+b)^n is of the form (nCk)·a^(n-k)·b^k for k =0 to n.