The Binomial Theorem

From Repeated Multiplication to a General Formula

A power such as $(x+y)^4$ can be expanded by multiplying four copies of $x+y$. As the exponent grows, however, collecting like terms by hand becomes increasingly cumbersome. The Binomial Theorem gives a single formula for every nonnegative integer exponent. Its coefficients, called binomial coefficients, record how often each type of term appears.

The proof will use only finite algebra: the distributive law, factorials, and induction on the exponent. A useful step is the relation between neighboring binomial coefficients known as Pascal’s identity. It explains how the coefficients in one expansion produce those in the next.

Factorials and Binomial Coefficients

For a positive integer $n$, the factorial $n!$ is the product of the positive integers from $1$ through $n$. We define $0!=1$. Thus $1!=1$, $2!=2\cdot1$, and $4!=4\cdot3\cdot2\cdot1=24$. For integers $n\geq0$ and $k$ with $0\leq k\leq n$, define the binomial coefficient as follows.

Definition (Binomial Coefficient). For integers $n\geq0$ and $0\leq k\leq n$, let $$ \binom{n}{k}=\frac{n!}{k!(n-k)!}. $$ In particular, $\binom{n}{0}=\binom{n}{n}=1$. The notation $\binom{n}{k}$ is read “$n$ choose $k$.”

The boundary values follow directly from the definition. For example, $\binom{n}{0}=n!/(0!n!)=1$, and $\binom{n}{n}=n!/(n!0!)=1$. The factorial formula also gives the symmetry $\binom{n}{k}=\binom{n}{n-k}$, since the denominator $k!(n-k)!$ is unchanged when the two factors are exchanged.

The next identity relates coefficients in adjacent rows. It is the algebraic rule that will make induction work.

Theorem (Pascal’s Identity). For $n\geq1$ and $1\leq k\leq n$, $$ \binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k}. $$ The boundary entries satisfy $\binom{n+1}{0}=\binom{n+1}{n+1}=1$.

Proof. For $1\leq k\leq n$, the factorial definition gives

\binom{n}{k}+\binom{n}{k-1} =\frac{n!}{k!(n-k)!}+\frac{n!}{(k-1)!(n-k+1)!}.

Use the common denominator $k!(n-k+1)!$. The first numerator becomes $n!(n-k+1)$, and the second becomes $n!k$. Therefore

\binom{n}{k}+\binom{n}{k-1} =\frac{n!(n-k+1)+n!k}{k!(n-k+1)!} =\frac{n!(n+1)}{k!(n-k+1)!} =\frac{(n+1)!}{k!((n+1)-k)!} =\binom{n+1}{k}.

The boundary entries are $1$ by the boundary calculation for binomial coefficients. This proves the stated identity and boundary values. $\square$

The same identity shows that every binomial coefficient is a nonnegative integer, even though its definition is written as a quotient. The row for $n=0$ consists only of $1$. Each later row has boundary entries $1$, and each interior entry is the sum of two entries in the preceding row by Pascal’s identity. Induction on the row number therefore shows that all entries are positive integers.

Worked Example: Building a Row of Coefficients

Starting with the row for $n=3$, whose entries are $1,3,3,1$, Pascal’s identity constructs the row for $n=4$. Keep the boundary entries $1$, and add neighboring entries in the previous row:

1,\qquad 1+3=4,\qquad 3+3=6,\qquad 3+1=4,\qquad 1.

Thus the entries for $n=4$ are $1,4,6,4,1$. The factorial definition confirms the interior values: $\binom{4}{1}=4!/(1!3!)=4$, $\binom{4}{2}=4!/(2!2!)=6$, and $\binom{4}{3}=4!/(3!1!)=4$.

A binomial is a sum of two terms, such as $x+y$ or $2a-b$. The theorem below gives the full expansion of any nonnegative integer power of such a sum. The case $n=0$ is included: by the convention for zeroth powers, $(x+y)^0=1$, and the formula has just one term.

Theorem (Binomial Theorem). For all $x,y\in\mathbb{R}$ and every integer $n\geq0$, $$ (x+y)^n=\sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^k. $$ Equivalently, $$ (x+y)^n =\binom{n}{0}x^n+\binom{n}{1}x^{n-1}y+\cdots +\binom{n}{n-1}xy^{n-1}+\binom{n}{n}y^n. $$

Proof. We use induction on $n$. When $n=0$, the sum has one term and gives

\sum_{k=0}^{0}\binom{0}{k}x^{0-k}y^k =\binom{0}{0}x^0y^0 =1 =(x+y)^0.

Now suppose the formula holds for some $n\geq0$. Multiplying both sides by $x+y$ gives

(x+y)^{n+1} =\left(\sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^k\right)(x+y).

Distribute $x+y$ across the finite sum, and separate the terms that result from multiplying by $x$ and by $y$:

(x+y)^{n+1} =\sum_{k=0}^{n}\binom{n}{k}x^{n+1-k}y^k +\sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^{k+1}.

In the second sum, replace the index $k$ by $j-1$. Its terms then have powers $x^{n+1-j}y^j$, with $1\leq j\leq n+1$. The first sum has that same form with $0\leq j\leq n$. The combined coefficient of $x^{n+1-j}y^j$, for $1\leq j\leq n$, is consequently

\binom{n}{j}+\binom{n}{j-1}=\binom{n+1}{j},

by Pascal’s identity. At the endpoints $j=0$ and $j=n+1$, there is just one term, with coefficient $1$. Hence

(x+y)^{n+1} =\sum_{j=0}^{n+1}\binom{n+1}{j}x^{n+1-j}y^j.

This proves the formula for $n+1$. By induction, it holds for every integer $n\geq0$. $\square$

Using the Expansion

The index $k$ in the theorem counts the power of $y$; at the same time, the power of $x$ is $n-k$. Thus every term has total exponent $n$. The coefficients are read from the corresponding row of binomial coefficients, and the signs are handled by treating a negative term as part of $y$.

Worked Example: Expanding a Binomial with a Negative Term

Expand $(2a-b)^4$. Apply the theorem with $x=2a$, $y=-b$, and $n=4$. The coefficients are $1,4,6,4,1$, so

$$ (2a-b)^4 =(2a)^4+4(2a)^3(-b)+6(2a)^2(-b)^2+4(2a)(-b)^3+(-b)^4. $$

Evaluating each term gives

(2a)^4=16a^4,\qquad 4(2a)^3(-b)=-32a^3b,\qquad 6(2a)^2(-b)^2=24a^2b^2,

4(2a)(-b)^3=-8ab^3,\qquad (-b)^4=b^4.

Therefore

$$ (2a-b)^4=16a^4-32a^3b+24a^2b^2-8ab^3+b^4. $$

The signs alternate here because the powers of $-b$ alternate in sign. In particular, the last term is positive because its exponent is even.

Worked Example: Expanding a Shifted Power

For $(x+3)^5$, take $y=3$. The coefficients in the row for $n=5$ are $1,5,10,10,5,1$. Substitution into the theorem gives

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

Since $5\cdot3=15$, $10\cdot9=90$, $10\cdot27=270$, $5\cdot81=405$, and $3^5=243$, the expansion is

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

The powers of $x$ decrease from $5$ to $0$, while the powers of $3$ increase from $0$ to $5$. This pattern is present in every binomial expansion.

Worked Example: Extracting One Coefficient

Find the coefficient of $u^2v^3$ in $(3u+2v)^5$. In the theorem’s general term, the power of $v$ is $k$. To obtain $v^3$, set $k=3$; then the power of $u$ is $5-3=2$, as required. The term is

\binom{5}{3}(3u)^2(2v)^3 =10\cdot9u^2\cdot8v^3 =720u^2v^3.

No other index produces $v^3$, because the power of $v$ in the term indexed by $k$ is exactly $k$. Therefore the requested coefficient is $720$.

Two Useful Sums of Binomial Coefficients

The theorem can produce identities about the coefficients themselves. Choosing particular values for $x$ and $y$ can make the left side simple, while the right side becomes a sum.

Corollary (Sum and Alternating Sum of the Coefficients). For every integer $n\geq0$, $$ \sum_{k=0}^{n}\binom{n}{k}=2^n. $$ For every integer $n\geq1$, $$ \sum_{k=0}^{n}(-1)^k\binom{n}{k}=0. $$

Proof. Set $x=1$ and $y=1$ in the Binomial Theorem. Since $1^{n-k}=1$ and $1^k=1$ for every index, it gives

2^n=(1+1)^n=\sum_{k=0}^{n}\binom{n}{k}.

For the second identity, set $x=1$ and $y=-1$. When $n\geq1$, the left side is $(1-1)^n=0^n=0$. On the right, each term is $\binom{n}{k}1^{n-k}(-1)^k=(-1)^k\binom{n}{k}$. Thus

0=(1-1)^n=\sum_{k=0}^{n}(-1)^k\binom{n}{k}.

Both identities follow directly from the theorem. The restriction $n\geq1$ in the alternating-sum identity matters: at $n=0$, the sum is $1$, not $0$. $\square$

Interpreting the Formula Correctly

The Binomial Theorem applies to nonnegative integer exponents. It does not, by itself, give an expansion formula for an arbitrary real exponent. Also, the coefficient $\binom{n}{k}$ belongs to the term with $y^k$ and $x^{n-k}$; reversing these powers without changing the index can lead to mistakes.

A reliable way to use the formula is to identify $x$, $y$, and $n$ first, then write the general term $\binom{n}{k}x^{n-k}y^k$. This makes the exponent pattern explicit and keeps negative signs attached to the correct power. For a single requested coefficient, match the required powers to $k$ and $n-k$ before doing numerical arithmetic.

Key takeaway. For a nonnegative integer $n$, the Binomial Theorem expands $(x+y)^n$ as a finite sum whose coefficients are $\binom{n}{k}$. Pascal’s identity explains how those coefficients fit together, and substitutions such as $x=y=1$ turn the expansion into useful coefficient identities.

Check Your Understanding

Use the factorial definition, Pascal’s identity, and the Binomial Theorem to answer these questions.

What are $\binom{7}{0}$ and $\binom{7}{7}$, and why do they have those values?
Use Pascal’s identity to find the entries in the row for $n=5$, starting from the row for $n=4$.
In the expansion of $(a+2b)^6$, which value of $k$ produces a term containing $b^4$? What is the corresponding power of $a$?
Why does the term involving $y^k$ in the Binomial Theorem have $x^{n-k}$?
Why is the alternating-sum identity stated for $n\geq1$, rather than $n\geq0$?

The Binomial Theorem

What You'll Learn

From Repeated Multiplication to a General Formula

Factorials and Binomial Coefficients

Worked Example: Building a Row of Coefficients