Pascal's Triangle - Binomial Expansions

Binomial Expansions

Pascal's triangle determines the coefficients which arise in binomial expansions. For an example, consider the expansion

(x + y)2 = x2 + 2xy + y2 = 1x2y0 + 2x1y1 + 1x0y2.

Notice the coefficients are the numbers in row two of Pascal's triangle: 1, 2, 1. In general, when a binomial like x + y is raised to a positive integer power we have:

(x + y)n = a₀xn + a₁xn−1y + a₂xn−2y2 + ... + a_n−1xyn−1 + a_nyn,

where the coefficients a_i in this expansion are precisely the numbers on row n of Pascal's triangle. In other words,

This is the binomial theorem.

Notice that the entire right diagonal of Pascal's triangle corresponds to the coefficient of yn in these binomial expansions, while the next diagonal corresponds to the coefficient of xyn−1 and so on.

To see how the binomial theorem relates to the simple construction of Pascal's triangle, consider the problem of calculating the coefficients of the expansion of (x + 1)n+1 in terms of the corresponding coefficients of (x + 1)n (setting y = 1 for simplicity). Suppose then that

Now

The two summations can be reorganized as follows:

$\begin{align} & \sum_{i=0}^{n } a_{i } x^{i+1} + \sum_{i=0}^n a_i x^i \\ & {} = \sum_{i=1}^{n+1} a_{i-1} x^{i } + \sum_{i=0}^n a_i x^i \\ & {} = \sum_{i=1}^{n } a_{i-1} x^{i } + \sum_{i=1}^n a_i x^i + a_0x^0 + a_{n}x^{n+1} \\ & {} = \sum_{i=1}^{n } (a_{i-1} + a_i)x^{i } + a_0x^0 + a_{n}x^{n+1} \\ & {} = \sum_{i=1}^{n } (a_{i-1} + a_i)x^{i } + x^0 + x^{n+1} \end{align}$

(because of how raising a polynomial to a power works, a₀ = a_n = 1).

We now have an expression for the polynomial (x + 1)n+1 in terms of the coefficients of (x + 1)n (these are the a_is), which is what we need if we want to express a line in terms of the line above it. Recall that all the terms in a diagonal going from the upper-left to the lower-right correspond to the same power of x, and that the a-terms are the coefficients of the polynomial (x + 1)n, and we are determining the coefficients of (x + 1)n+1. Now, for any given i not 0 or n + 1, the coefficient of the xi term in the polynomial (x + 1)n+1 is equal to a_i (the figure above and to the left of the figure to be determined, since it is on the same diagonal) + a_i−1 (the figure to the immediate right of the first figure). This is indeed the simple rule for constructing Pascal's triangle row-by-row.

It is not difficult to turn this argument into a proof (by mathematical induction) of the binomial theorem. Since (a + b)n = bn(a/b + 1)n, the coefficients are identical in the expansion of the general case.

An interesting consequence of the binomial theorem is obtained by setting both variables x and y equal to one. In this case, we know that (1 + 1)n = 2n, and so

In other words, the sum of the entries in the nth row of Pascal's triangle is the nth power of 2.

Read more about this topic: Pascal's Triangle