In the Algebra course we have learned to expand the expression \((x+y)^2\) to \(x^2 + 2xy + y^2\text{.}\) In this Section we expand \((x+y)^n\) for arbitrary nonnegative integers \(n\text{.}\) For this, let us first recall how such an expression should be calculated.
Consider first \((x+y)^2\text{.}\) This is the product of \((x+y)\) by itself, that is, \((x+y)^2 = (x+y)\cdot (x+y)\text{,}\) and has to be computed by multiplying every term of the first factor by every term of the second factor:
\begin{equation} (x+y)^2 = (x+y) \cdot (x+y) = x^2 + xy + yx + y^2 = x^2 + 2xy + y^2\text{.}\label{eq__x_y__2}\tag{4.1.1} \end{equation}Now, consider \((x+y)^3\text{.}\) This is the threefold product of \((x+y)\) with itself, that is, \((x+y)^3 = (x+y)\cdot (x+y) \cdot (x+y)\text{.}\) Note, that this is the same as the product \((x+y)^2 \cdot (x+y)\text{,}\) for which the first factor we have already computed in (4.1.1).
\begin{align*} (x+y)^3 \amp = (x+y)^2 \cdot (x+y) = (x^2 + 2xy + y^2)\cdot (x+y)\\ \amp = x^3 + x^2y + 2x^2y + 2xy^2 + xy^2 + y^3\\ \amp = x^3 + 3x^2y + 3 xy^2 + y^3\text{.} \end{align*}This way, we can easily continue calculating the higher powers of \((x+y)\text{:}\)
\begin{align*} (x+y)^4 \amp = (x+y)^3 \cdot (x+y) = (x^3 + 3x^2y + 3 xy^2 + y^3)\cdot (x+y)\\ \amp = x^4 + x^3y + 3x^3y + 3x^2y^2 + 3x^2y^2 + 3xy^3 + xy^3 + y^4\\ \amp = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4.\\ (x+y)^5 \amp = (x+y)^4 \cdot (x+y) = (x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4)\cdot (x+y)\\ \amp = x^5 + x^4y + 4x^4y + 4x^3y^2 + 6x^3y^2 + 6x^2y^3 + 4x^2y^3 + 4xy^4 + xy^4 + y^5\\ \amp = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5.\\ (x+y)^6 \amp = (x+y)^5 \cdot (x+y) = (x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5)\cdot (x+y)\\ \amp = x^6 + x^5y + 5x^5y + 5x^4y^2 + 10x^4y^2 + 10x^3y^3 + 10x^3y^3 + 10x^2y^4\\ \amp + 5x^2y^4 + 5xy^5 + xy^5 + y^6\\ \amp = x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6\text{.} \end{align*}Let us summarize our findings:
\begin{align*} (x+y)^2 \amp = x^2 + 2xy + y^2,\\ (x+y)^3 \amp = x^3 + 3x^2y + 3 xy^2 + y^3,\\ (x+y)^4 \amp = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4,\\ (x+y)^5 \amp = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5,\\ (x+y)^6 \amp = x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6\text{.} \end{align*}Now, wait a minute! The coefficients arising in these expressions are exactly the numbers occurring in Pascal's triangle. Indeed, the coefficients of \((x+y)^6\) are \(1 = \binom{6}{0}\text{,}\) \(6 = \binom{6}{1}\text{,}\) \(15 = \binom{6}{2}\text{,}\) \(20 = \binom{6}{3}\text{,}\) \(15 = \binom{6}{4}\text{,}\) \(6 = \binom{6}{5}\text{,}\) \(1 = \binom{6}{6}\text{.}\) This cannot be a coincidence! It looks like that when we expand \((x+y)^n\text{,}\) then the coefficient for the term \(x^{n-k}y^k\) is \(\binom{n}{k}\text{.}\) This is always the case, not only for the first six powers. This is the statement of the binomial theorem.
Let \(n\) be a natural number. Then
\begin{align*} (x+y)^n \amp = x^n + n x^{n-1}y + \binom{n}{2} x^{n-2}y^2 + \dots + \binom{n}{n-2} x^2 y^{n-2} + n x y^{n-1} + y^n\\ \amp = \sum_{k=0}^n \binom{n}{k} x^{n-k}y^k\text{.} \end{align*}Note first, that the Binomial theorem holds for \(n=0\) and \(n=1\text{,}\) as well: \((x+y)^0 = 1 = \binom{0}{0} x^0 y^0\text{,}\) \((x+y)^1 = x + y = \binom{1}{0} x^1 y^0 + \binom{1}{1} x^0 y^1\text{.}\) Now, we can prove the theorem by induction on \(n\text{.}\) Assume that the statement holds for \(n-1\text{,}\) that is,
\begin{equation*} (x+y)^{n-1} = \sum_{k=0}^{n-1} \binom{n-1}{k} x^{n-1-k}y^k\text{.} \end{equation*}This is the induction hypothesis. Now, compute \((x+y)^n\) using the same method as before, and use the induction hypothesis for expanding \((x+y)^{n-1}\text{:}\)
\begin{align*} (x+y)^n \amp = (x+y)^{n-1} \cdot (x+y) = \left( \sum_{k=0}^{n-1} \binom{n-1}{k} x^{n-1-k}y^k \right) \cdot \left( x + y \right)\\ \amp = \sum_{k=0}^{n-1} \binom{n-1}{k} x^{n-1-k}y^k \cdot x + \sum_{k=0}^{n-1} \binom{n-1}{k} x^{n-1-k}y^k \cdot y\\ \amp = \sum_{k=0}^{n-1} \binom{n-1}{k} x^{n-k}y^k + \sum_{k=0}^{n-1} \binom{n-1}{k} x^{n-1-k}y^{k+1}\\ \amp = x^n + \sum_{k=1}^{n-1} \binom{n-1}{k} x^{n-k}y^k + \sum_{k=0}^{n-2} \binom{n-1}{k} x^{n-1-k}y^{k+1} + y^n\\ \amp = x^n + \sum_{k=1}^{n-1} \binom{n-1}{k} x^{n-k}y^k + \sum_{k=1}^{n-1} \binom{n-1}{k-1} x^{n-k}y^k + y^n\\ \amp = x^n + \sum_{k=1}^{n-1} \left( \binom{n-1}{k} + \binom{n-1}{k-1}\right) x^{n-k}y^k + y^n\\ \amp = x^n + \sum_{k=1}^{n-1} \binom{n}{k} x^{n-k}y^k + y^n = \sum_{k=0}^n \binom{n}{k} x^{n-k}y^k\text{.} \end{align*}Here, we have separated \(x^n\) and \(y^n\) from the sums in <<Unresolved xref, reference "eq_x_nfront"; check spelling or use "provisional" attribute>> , then “re-indexed” the second sum in <<Unresolved xref, reference "eq_binomreindex"; check spelling or use "provisional" attribute>> to find the coefficient of the common terms \(x^{n-k}y^k\) (for \(k = 1, 2, \dots , n-1\)) of the two sums. Finally, in <<Unresolved xref, reference "eq_binomsum"; check spelling or use "provisional" attribute>> we used the generating rule of Pascal's triangle (Proposition 4.0.3).
Repeat the proof by “writing out” all sums.
Now we understand why binomial coefficients are called like that: because they arise as the coefficients in the \(n\)th power of binomial sums. Moreover, the proof of the Binomial theorem revealed that raising \((x+y)\) to the next power affects the coefficients exactly the same way as we generate Pascal's triangle. Nevertheless, one can find another argument, which explains “better” why the binomial coefficients arise in the \(n\)th power.
Consider \((x+y)^6\text{:}\)
\begin{equation*} (x+y)^6 = x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6\text{.} \end{equation*}How do we obtain the coefficient 15 for \(x^4y^2\text{?}\) Now, \((x+y)^6\) is the 6-fold product of \((x+y)\) by itself:
\begin{equation*} (x+y)^6 = (x+y) \cdot (x+y) \cdot (x+y) \cdot (x+y) \cdot (x+y) \cdot (x+y)\text{.} \end{equation*}The multiplication of these six factors is carried out by choosing a term from each factor (\(x\) or \(y\)) in every possible way, multiplying these six terms, and then adding the resulting products together. Thus the coefficient of \(x^4y^2\) is the number of possibilities to choose four times the \(x\) and two times the \(y\) out of the six factors. Altogether there are six \(y\)'s to choose from, and we need to choose two of them (and the remaining four factors will be chosen as \(x\)). This can be done in \(\binom{6}{2} = 15\)-many ways. Therefore the coefficient of \(x^4y^2\) is \(\binom{6}{2} = 15\text{.}\)
Prove the Binomial Theorem using the argument provided above.
The Binomial theorem can be used to calculate several \(n\)th powers. For example, choosing \(y=1\text{,}\) every power of \(y\) is 1, as well, thus
\begin{align*} (x+1)^n \amp = x^n + n x^{n-1}\cdot 1 + \binom{n}{2} x^{n-2}\cdot 1^2 + \dots + n x \cdot 1^{n-1} + 1^n\\ \amp = x^n + n x^{n-1} + \binom{n}{2} x^{n-2} + \dots + n x + 1 = \sum_{k=0}^n \binom{n}{k} x^k\text{.} \end{align*}Write \(x=y=1\) into the Binomial theorem. Note that this provides a second proof for Proposition 2.6.9.
Alternatively, we can substitute \(-y\) instead of \(y\) in the Binomial theorem, obtaining
\begin{align*} (x-y)^n \amp = x^n + n x^{n-1}\cdot (-y) + \binom{n}{2} x^{n-2}\cdot (-y)^2 + \dots + n x \cdot (-y)^{n-1} + (-y)^n\\ \amp = x^n - n x^{n-1} y + \binom{n}{2} x^{n-2} y^2 - \dots + (-1)^{n-1} n x y^{n-1} + (-1)^n y^n\\ \amp = \sum_{k=0}^n (-1)^k \binom{n}{k} x^{n-k}y^{k}\text{.} \end{align*}Choosing \(y=-1\) yields
\begin{align*} (x-1)^n \amp = x^n + n x^{n-1}\cdot (-1) + \binom{n}{2} x^{n-2}\cdot (-1)^2 + \dots + n x \cdot (-1)^{n-1} + (-1)^n\\ \amp = x^n - n x^{n-1} + \binom{n}{2} x^{n-2} - \dots + (-1)^{n-1} n x + (-1)^n\\ \amp = \sum_{k=0}^n (-1)^k \binom{n}{k} x^{n-k}\text{.} \end{align*}Write \(x=1\text{,}\) \(y=-1\) into the Binomial theorem. What do you observe?
Expand the following expressions: \((x+y)^8\text{,}\) \((x-y)^8\text{,}\) \((a+1)^{10}\text{,}\) \((b-3)^5\text{,}\) \((1 + 2/x)^5\text{,}\) \(\left( a + b \right)^6\text{,}\) \(\left( 1 + x \right)^5\text{,}\) \(\left(3a + 4b \right)^4\text{,}\) \(\left( 3 - 2x \right)^4\text{.}\)
In the binomial expansion of \(\left( 1- x/2 \right)^9\text{,}\) written in terms of ascending powers of \(x\text{,}\) find the fourth term. Then find the coefficient of \(x^5\text{.}\)