The binomial theorem and Pascal’s triangle

2026-10-22 huynhvietkhanh Comments: 0

Derive the binomial theorem by counting, construct Pascal’s triangle, and find coefficients and specified terms.

High schoolCounting & Probability

Content updated: October 22, 2026

The binomial theorem collects the products obtained by multiplying identical factors. A binomial coefficient counts the choices of positions contributing the second term, connecting the expansion directly to counting.

Learning goals: expand a binomial, construct Pascal’s triangle, and find the term containing a specified power.

Prerequisites: combinations and Pascal’s identity (English lesson); powers and polynomial multiplication.

On this page: The idea · Method · Examples · Practice · Further learning

1. The binomial formula and why combinations occur

Two familiar expansions are

$$\begin{aligned}(a+b)^2&=a^2+2ab+b^2,\\(a+b)^3&=a^3+3a^2b+3ab^2+b^3.\end{aligned}$$

The distributive law also produces higher powers, such as degree $7$, but the number of products to collect grows rapidly. For an integer $n\ge1$ and real or complex numbers $a,b$, the general formula is

$$(a+b)^n=\sum_{k=0}^nC_n^k a^{n-k}b^k.$$

Indeed, write $(a+b)^n$ as a product of $n$ factors $(a+b)$. Each distributed product selects $a$ or $b$ from every factor. To obtain $a^{n-k}b^k$, choose exactly $k$ positions contributing $b$, in $C_n^k$ ways. Summing over $k$ from $0$ to $n$ establishes the formula. Commutativity permits collecting products with the same numbers of occurrences of $a,b$. For $n=0$, both sides are interpreted as the constant polynomial $1$.

2. Symmetry, Pascal’s triangle, and the general term

Since $C_n^k=C_n^{n-k}$, the coefficient sequence is symmetric from its two ends. Both boundary coefficients equal $1$. Construct successive rows with Pascal’s identity $C_n^k+C_n^{k+1}=C_{n+1}^{k+1}$: each interior entry is the sum of the two entries immediately above it.

$$\begin{array}{c|rrrrrr}n&&&&&&\\\hline0&1&&&&&\\1&1&1&&&&\\2&1&2&1&&&\\3&1&3&3&1&&\\4&1&4&6&4&1&\\5&1&5&10&10&5&1\end{array}$$

For instance, $C_1^0+C_1^1=C_2^1=2$ and $C_2^1+C_2^2=C_3^2=3$. Rows $n=4,5$ continue by the same rule, without requiring a separate diagram. Row $n$ has $n+1$ coefficient positions.

When the powers of $b$ are written in increasing order, term $k+1$ is

$$T_{k+1}=C_n^k a^{n-k}b^k,\qquad k\in\{0,1,\ldots,n\}.$$

To find the coefficient of $x^r$, substitute the specified expressions for $a,b$, simplify the exponent of $x$, solve the exponent equation, and check that $k$ is an integer in $[0;n]$. Then calculate the coefficient, distinguishing it from the full term containing $x^r$.

3. Worked examples

Example 1. A seventh-power expansion

The coefficients $C_7^k$, in the order $k=0,1,\ldots,7$, form the sequence $1,7,21,35,35,21,7,1$. Hence

$$\begin{aligned}(a+b)^7={}&a^7+7a^6b+21a^5b^2+35a^4b^3\\&+35a^3b^4+21a^2b^5+7ab^6+b^7.\end{aligned}$$

In each term, the exponents of $a,b$ sum to $7$. The coefficient sum is $2^7=128$, as checked by substituting $a=b=1$.

Example 2. Symmetric coefficients

In the preceding expansion, the first and last coefficients are $1,1$, followed by $7,7$ and $21,21$, with both middle coefficients equal to $35$. These are the identities $C_7^0=C_7^7$, $C_7^1=C_7^6$, $C_7^2=C_7^5$, and $C_7^3=C_7^4$. The symmetry concerns the coefficient sequence; it does not assert equality of the terms for all $a,b$.

Example 3. Finding the term containing x to the eighth power

Find the term containing $x^8$ in $(x^2+2)^{10}$. The general term is

$$T_{k+1}=C_{10}^k(x^2)^{10-k}2^k=C_{10}^k2^k x^{20-2k}.$$

The condition $20-2k=8$ is equivalent to $k=6$, a valid integer index. Thus the required term is

$$C_{10}^6\,2^6x^8=210\cdot64x^8=13440x^8.$$

The coefficient of $x^8$ is $13440$; the full term is $13440x^8$. Since $k=6$, it is term $7$ in the specified ordering, not term $6$.

4. Practice exercises

Find the coefficient of $x^3$ in $(1-2x)^5$.
Find the term independent of $x$ in $(2x-1/x)^6$, where $x\ne0$.
Does $(x^2+2)^{10}$ have a term containing $x^7$? Explain using the index $k$.

Show answers and hints

With $a=1,b=-2x$, choose $k=3$: the coefficient is $C_5^3(-2)^3=10(-8)=-80$.
The general term is $C_6^k2^{6-k}(-1)^kx^{6-2k}$. Its exponent equals $0$ when $k=3$, so the term independent of $x$ is $C_6^3\cdot8\cdot(-1)=-160$.
No: $20-2k=7$ gives $k=13/2$, which is not an integer. Equivalently, every exponent in the expansion is even.

5. Common mistakes

Applying this finite expansion to an exponent other than a nonnegative integer without another theorem.
Losing the sign of $b$ when $b=-2x$, or raising only $x$ to the power and omitting its coefficient.
Accepting a value of $k$ that is not an integer or lies outside $[0;n]$.
Confusing the index $k$ with the term position $k+1$, or a coefficient with the full term.

6. Related lessons

Original exercise resource on Scribd.