The Binomial Theorem: Free Response

Row 9 has two easy entries at its ends and eight interior entries built one addition at a time from row 8.

Start adjacent to a boundary value and add across row 8:

$\binom{9}{1}=\binom{8}{0}+\binom{8}{1}=1+8=9$

Continue the same way across the rest of the row:

$\binom{9}{2}=8+28=36,\quad \binom{9}{3}=28+56=84,\quad \binom{9}{4}=56+70=126$

$\binom{9}{5}=70+56=126,\quad \binom{9}{6}=56+28=84,\quad \binom{9}{7}=28+8=36,\quad \binom{9}{8}=8+1=9$

Putting the boundary values back at the two ends gives the complete row.

$1,\ 9,\ 36,\ 84,\ 126,\ 126,\ 84,\ 36,\ 9,\ 1$

Ten entries in all, one for each value of $k$ from $0$ to $9$, matching the $n+1$ entries every row of the triangle has.

The definition of $\binom{9}{k}$ does not depend on which row it sits in: it is the number of ways to choose which $k$ of the $9$ factors of $(x+y)^9$ donate a $y$ instead of an $x$. Setting $k=4$, $\binom{9}{4}$ counts the ways to choose which $4$ of the $9$ factors donate a $y$; the remaining $5$ factors then donate an $x$ automatically.

That is exactly the situation the term $x^5y^4$ describes: $5$ factors gave up an $x$ and $4$ gave up a $y$. So the coefficient of that term is nothing but this count.

$\text{coefficient of } x^5y^4 = \binom{9}{4} = 126$

A choice of which $4$ of the $9$ factors donate a $y$ automatically decides which $5$ donate an $x$: fix one and the other is forced, since every factor donates exactly one letter. So the two descriptions, choosing $4$ factors to donate $y$ and choosing $5$ factors to donate $x$, name the very same set of choices, described from two directions.

Two counts of the same set of choices have to be equal, whatever that set turns out to be.

$\binom{9}{4} = \binom{9}{5}$

Row $9$ confirms it directly: the fifth and sixth entries, sitting on either side of the row's middle, both read $126$.

Expand using the theorem with $A=x$, $B=3$, and $n=5$, taking the coefficients from row $5$: $1, 5, 10, 10, 5, 1$.

$(x+3)^5 = x^5 + 5x^4(3) + 10x^3(3)^2 + 10x^2(3)^3 + 5x(3)^4 + (3)^5$

Evaluate each power of $3$ and combine it with its coefficient:

$5(3)=15,\quad 10(9)=90,\quad 10(27)=270,\quad 5(81)=405,\quad 3^5=243$

So the expansion is

$(x+3)^5 = x^5 + 15x^4 + 90x^3 + 270x^2 + 405x + 243.$

Write $x-3$ as $x+(-3)$ and let the same row-5 coefficients apply to $A=x$, $B=-3$.

$(x-3)^5 = x^5 + 5x^4(-3) + 10x^3(-3)^2 + 10x^2(-3)^3 + 5x(-3)^4 + (-3)^5$

Every magnitude here is exactly the one already computed in part A, since $(-3)^k$ has the same size as $3^k$; only the sign of each power of $-3$ is new.

$(-3)^1=-3,\ (-3)^2=9,\ (-3)^3=-27,\ (-3)^4=81,\ (-3)^5=-243$

The odd powers come out negative and the even powers come out positive, giving

$(x-3)^5 = x^5 - 15x^4 + 90x^3 - 270x^2 + 405x - 243.$

Line the two expansions up term by term.

$x^5+15x^4+90x^3+270x^2+405x+243 \quad\text{and}\quad x^5-15x^4+90x^3-270x^2+405x-243$

The terms at $k=0,2,4$, namely $x^5$, $90x^3$, and $405x$, are identical in both expansions. The terms at $k=1,3,5$, namely $15x^4$, $270x^2$, and $243$, have opposite signs in the two expansions.

The reason traces straight back to $(-3)^k=(-1)^k3^k$. Whenever $k$ is even, $(-1)^k=1$ and the term is unaffected by the change from $+3$ to $-3$. Whenever $k$ is odd, $(-1)^k=-1$ and the term's sign reverses. Nothing about the size of any term changes; only the terms at odd $k$ have anything to flip.

Apply the general term formula $T_k=\binom{n}{k}A^{n-k}B^k$ with $A=x$, $B=3y$, and $n=12$.

$T_k = \binom{12}{k}x^{12-k}(3y)^k$

The factor $(3y)^k$ raises the whole term $3y$ to the power $k$, so split it into $3^k$ and $y^k$ before the term is in its simplest form.

$T_k = \binom{12}{k}3^kx^{12-k}y^k$

The term containing $x^9y^3$ needs the exponent on $x$ to equal $9$.

$12-k=9 \quad\Longrightarrow\quad k=3$

That same $k$ automatically gives $y^3$, matching the target, so this is the right term to evaluate. Substitute $k=3$ into the general term from part A.

$T_3 = \binom{12}{3}3^3x^9y^3 = 220 \cdot 27 \cdot x^9y^3 = 5940\,x^9y^3$

So the coefficient is $5940$.

The list of terms runs $T_0, T_1, T_2, \ldots, T_{12}$, starting the count at $k=0$ rather than at $k=1$. So $T_k$ has exactly $k$ terms sitting in front of it in that list, at $k=0$ through $k=k-1$, which makes it the $(k+1)$st term overall.

$\text{term number} = k+1$

Part B found $k=3$ for the term containing $x^9y^3$, so its term number is $3+1=4$: the fourth term, not the third. The classmate's mistake was reading the exponent on $y$ as if the list of terms started counting at $1$ instead of $0$; that off-by-one gap between $k$ and the term number is fixed, and it applies to every expansion, not only this one.

A quotient with $x$ in the denominator becomes a negative exponent:

$\frac{2}{x} = 2x^{-1}$

Apply the general term formula with $A=x^3$, $B=2x^{-1}$, and $n=7$.

$T_k = \binom{7}{k}(x^3)^{7-k}(2x^{-1})^k = \binom{7}{k}2^kx^{3(7-k)}x^{-k}$

Multiply the two powers of $x$ by adding their exponents, $3(7-k)-k=21-3k-k=21-4k$.

$T_k = \binom{7}{k}2^kx^{21-4k}$

Set the exponent from part A equal to the target power.

$21-4k=9 \quad\Longrightarrow\quad 4k=12 \quad\Longrightarrow\quad k=3$

Substitute $k=3$ into the general term.

$T_3 = \binom{7}{3}2^3x^9 = 35 \cdot 8 \cdot x^9 = 280\,x^9$

So the coefficient is $280$.

A constant term is the term whose exponent on $x$ is $0$, so set the exponent from part A equal to $0$ and solve.

$21-4k=0 \quad\Longrightarrow\quad k=\frac{21}{4}$

That is not a whole number. But $k$ can only ever be a whole number from $0$ to $n=7$, since it counts how many of the $7$ factors donated the $2x^{-1}$ piece, and there is no such thing as a fraction of a factor. Because no whole number satisfies the equation, no value of $k$ in the allowed range produces a constant term, so none of the eight terms of this expansion is constant.

Write out $\binom{n}{1}$ using the closed form and cancel $(n-1)!$ against the tail of $n!$.

$\binom{n}{1} = \frac{n!}{1!(n-1)!} = \frac{n \cdot (n-1)!}{(n-1)!} = n$

Do the same for $\binom{n}{2}$, canceling $(n-2)!$ against the tail of $n!$.

$\binom{n}{2} = \frac{n!}{2!(n-2)!} = \frac{n(n-1)(n-2)!}{2 \cdot (n-2)!} = \frac{n(n-1)}{2}$

Substitute the two expressions from part A into the given equation.

$\frac{n(n-1)}{2} = 5n$

Multiply both sides by $2$ to clear the fraction, then move everything to one side and factor.

$n(n-1) = 10n \quad\Longrightarrow\quad n^2 - 11n = 0 \quad\Longrightarrow\quad n(n-11)=0$

A product is zero only when one of its factors is, so $n=0$ or $n=11$. The problem requires $n \ge 2$, so $n=0$ is discarded, leaving

$n=11$

Check by evaluating both sides of the original equation directly: $\binom{11}{2}=\dfrac{11 \cdot 10}{2}=55$, and $5\binom{11}{1}=5(11)=55$. The two agree.

Pascal's rule builds one row from the row before it, always moving forward and never backward: to reach row $11$ you would have to build every row from $0$ up to it in order, and the only way to know you have reached the right row is to already suspect the answer is $11$. Nothing in the rule itself points to which row satisfies $\binom{n}{2}=5\binom{n}{1}$; you would be searching, not solving.

The closed-form formula sidesteps that entirely, because $n$ appears as a genuine algebraic variable in both expressions from part A.

$\binom{n}{1}=n, \qquad \binom{n}{2}=\frac{n(n-1)}{2}$

That is what allowed part B to set up and solve an equation in $n$ directly, the same way any other algebraic equation is solved, rather than checking row after row by hand.