Binomial Theorem¶

Theorem¶

For a nonnegative integer $n$ and real numbers $a$ and $b$,

Interpretation¶

Expanding the power $(a+b)^n$ means multiplying $n$ factors $a+b$ and choosing $a$ or $b$ from each factor. A product in which $a$ is chosen $i$ times equals $a^ib^{n-i}$, and the binomial coefficient $\binom{n}{i}$ counts the ways to pick those $i$ factors.

Row $n$ of Pascal's triangle lists $\binom{n}{0},\ldots,\binom{n}{n}$. In the highlighted row $n=3$$3$, the coefficients $1,3,3,1$ belong to $a^3$, $a^2b$, $ab^2$ and $b^3$: the term $a^2b$ appears $\binom{3}{2}=3$ times, once for each choice of the two factors that supply $a$.

In probability the theorem is used with $a=p$ and $b=1-p$$p$. The terms $\binom{n}{i}p^i(1-p)^{n-i}$ are then the binomial probabilities of $i$ successes in $n$ independent trials, and the theorem shows that they add up to $(p+1-p)^n=1$$1$. Weighting the $i$th term by $\mathrm{e}^{ti}$ replaces $p$ by $p\mathrm{e}^t$, so the same step gives the MGF $(1-p+p\mathrm{e}^t)^n$ of $\operatorname{Bin}(n,p)$. With $a=b=1$ the theorem counts subsets: $\displaystyle \sum_{i=0}^n\binom{n}{i}=2^n$$2^n$.

A common slip is to expand $(a+b)^n$ as $a^n+b^n$$b^n$; already for $n=2$ this misses the cross term $2ab$.

Key Results¶

• Setting $a=p$$p$, $b=1-p$ gives $\displaystyle \sum_{k=0}^n\binom nkp^k(1-p)^{n-k}=1$$1$.
• The coefficient of $a^kb^{n-k}$ is $\binom nk=n!/[k!(n-k)!]$$k)!]$.

Examples¶

• $(a+b)^3=a^3+3a^2b+3ab^2+b^3$$b^3$.
• For two independent fair tosses, the probabilities of zero, one, and two heads are $1/4,1/2,1/4$, the terms in $(1/2+1/2)^2$$1/2{)}^2$.

Prerequisites¶

• Binomial Coefficient

Related Concepts¶

• Binomial Distribution
• Multinomial Distribution

Sources¶

• Nicky van Foreest · Probability Distributions · Refresher

Aliases: binomial expansion Symbols: sum

Last updated: 20 October 2026 03:02