Binomial Theorem

Summary: The formula expressing $(a+b{)}^n$ as a sum of terms with binomial coefficients; valid for all integer $n$ as a finite sum and extended to fractional and negative $n$ as an infinite series.

Sources: chapter-1.2.10, chapter-1.2.11, chapter-1.2.12, chapter-1.2.13

Last updated: 2026-04-29

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Integer Exponents (finite form)

For a non-negative integer $n$:

$$(a+b{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a^{n-k}{b}^k$$

where the binomial coefficient is

$$\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n(n-1)(n-2)\cdots (n-k+1)}{k!}=\frac{n!}{k!(n-k)!}$$

Each term $\left(\genfrac{}{}{0ex}{}{n}{k}\right)a^{n-k}{b}^k$ has exponents summing to $n$. The coefficients form row $n$ of pascal-triangle. (source: chapter-1.2.10)

For $(a-b{)}^n$, the same formula applies with the signs of odd-position terms negated. (source: chapter-1.2.10)

Coefficient Formula (Euler’s fractions)

Euler writes the coefficients of the $n$th power using the fraction sequence $\frac{n}{1},\frac{n-1}{2},\dots ,\frac{1}{n}$:

• 1st coefficient = 1
• 2nd coefficient = $\frac{n}{1}$
• $k$th coefficient = product of the first $k-1$ fractions

This is numerically identical to $\left(\genfrac{}{}{0ex}{}{n}{k-1}\right)$ and allows any coefficient to be computed without Pascal’s triangle. (source: chapter-1.2.10)

Combinatorial Proof

The coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ counts the number of distinct orderings of $k$ copies of $b$ and $n-k$ copies of $a$, which equals $n!/(k!(n-k)!)$. Proved in ch1.2.11-transpositions-and-binomial-proof. (source: chapter-1.2.11)

General Exponent (infinite series)

The same coefficient formula extends to any real $n$. When $n$ is not a non-negative integer, the series does not terminate:

$$(a+b{)}^n=a^n+\frac{n}{1}{a}^{n-1}b+\frac{n(n-1)}{1\cdot 2}{a}^{n-2}{b}^2+\frac{n(n-1)(n-2)}{1\cdot 2\cdot 3}{a}^{n-3}{b}^3+\cdots$$

Key cases treated by Euler:

$n$	Expression	Chapter
$\frac{1}{2}$	$\sqrt{a+b}$ as a series	ch1.2.12-irrational-powers-infinite-series
$\frac{1}{3}$	$\sqrt[3]{a+b}$ as a series	ch1.2.12-irrational-powers-infinite-series
$-1$	$1/(a+b)$ — geometric series	ch1.2.13-negative-powers-infinite-series
$-2$	$1/(a+b{)}^2$ — natural-number coefficients	ch1.2.13-negative-powers-infinite-series
$-3$	$1/(a+b{)}^3$ — triangular numbers	ch1.2.13-negative-powers-infinite-series

(sources: chapter-1.2.12, chapter-1.2.13)

Verification

Setting $a=b=1$ gives $(1+1{)}^n=2^n$, which must equal the sum of all coefficients in the expansion. Euler verifies this for rows 1–7 of the integer table. (source: chapter-1.2.10)

For the infinite-series cases, Euler verifies by multiplying the series back by the appropriate power of $(a+b)$ and checking that all intermediate terms cancel, leaving 1. (source: chapter-1.2.13)

Historical Note

Euler presents what is now called Newton’s generalised binomial series, observing that the coefficient formula found for positive integers extends naturally to fractions and negatives. The convergence condition (that the ratio $b/a$ be small) is used implicitly through the iterative recentering procedure, but not stated as a formal criterion. (source: chapter-1.2.12)

Related pages

• pascal-triangle
• factorials
• infinite-series
• ch1.2.10-higher-powers-compound
• ch1.2.11-transpositions-and-binomial-proof
• ch1.2.12-irrational-powers-infinite-series
• ch1.2.13-negative-powers-infinite-series
• algebraic-identities
• powers-and-exponents