Binomial Theorem

Derive the coefficients by counting choices

For a positive integer $n$, the product $(a+b)^n$ contains $n$ copies of $(a+b)$. Each term in the expanded product chooses either $a$ or $b$ from every copy.

To obtain $a^{n-k}b^k$, choose $b$ from exactly $k$ of the $n$ factors. There are $\binom nk$ such selections. Collecting equal terms gives

where $\binom nk=n!/[k!(n-k)!]$ for integers $0\le k\le n$.

For $n=0$, the polynomial expansion is the constant 1. Treat the formula as a polynomial identity: a factor with exponent zero contributes 1. This avoids interpreting a displayed endpoint term as an unrelated numerical question about $0^0$.

The proof counts selections, not probabilities. It is valid for real or complex $a,b$ and does not require $a+b$ to be positive. The OpenStax binomial-theorem chapter gives the standard expansion and coefficient methods.

Use Pascal’s rule to organise the coefficients

The first rows of binomial coefficients are

For $1\le k\le n-1$, Pascal’s rule is

To count $k$-element selections from $n$ objects, separate those containing a chosen object from those not containing it. The two cases are disjoint and exhaust all selections, which proves the identity.

The edge coefficients are 1 because there is one way to select none or all of the objects. Symmetry $\binom nk=\binom n{n-k}$ follows by matching each selection with its complementary selection.

Expand a signed binomial carefully

For $(2x-3)^4$, use $a=2x$ and $b=-3$, retaining the negative sign inside each power:

Simplification gives

The alternating signs come from the powers of $-3$. They are not supplied by changing the coefficients in Pascal’s triangle. A constant-term check gives $(-3)^4=81$, and a leading-term check gives $(2x)^4=16x^4$.

At $x=1$, the coefficients sum to $16-96+216-216+81=1$, matching $(2-3)^4$. These checks are useful, but the term-by-term derivation establishes the entire polynomial identity.

Find one coefficient without expanding everything

Find the coefficient of $x^5$ in $(2x-3)^8$. Its general term, indexed by $k$, is

The power of $x$ is $8-k$. Set $8-k=5$ to obtain $k=3$. Then the required coefficient is

This is the fourth term in the conventional descending-power order, because its index is $k+1=4$. The value $k=3$ counts selections of the second binomial term; it is not the ordinal position of the term.

Solve the exponent equation for a constant or missing term

Consider

The general term is

A constant term requires $12-3k=0$, so $k=4$. Its value is $\binom64=15$.

Now ask for an $x^5$ term in $(x^2+1/x)^8$. The exponent equation is $16-3k=5$, which gives $k=11/3$. That is not an integer index from 0 through 8. Therefore the coefficient of $x^5$ is zero: no such term occurs.

The presence of reciprocals makes these Laurent-polynomial expressions rather than ordinary polynomials in $x$, but the binomial expansion remains a finite identity on $x\ne0$. Preserve that original restriction.

Use an exact expansion to assess an approximation

For $1.02^5$, write $1.02=1+u$ with $u=0.02$. The finite identity gives

Substituting $u=0.02$ gives

The first-order approximation $1+5u=1.1$ is below the exact result by 0.0040808032. Its direction is known because every omitted term is positive for this positive $u$.

Keeping the quadratic term gives 1.104, with error 0.0000808032. Calling either expression an approximation rather than an equality preserves the distinction between truncating an expansion and using the full finite identity.

For negative $u$, omitted terms can have different signs. Do not carry over a positive-term error argument unchanged. For a noninteger exponent, there is generally no terminating expansion, so an error bound requires a different argument.

Make the coefficient checks meaningful

For $P(x)=(2x-3)^4$, the sum of its coefficients is $P(1)=1$. The alternating sum, obtained by substituting $x=-1$, is $P(-1)=625$. The coefficient magnitudes also sum to 625 here because the signs alternate in descending powers for this even exponent.

More generally, the sum of all binomial coefficients in row $n$ is $2^n$, obtained by setting $a=b=1$. For $n\ge1$, their alternating sum is zero, obtained by setting $a=1$, $b=-1$.

These are evaluations of an identity with well-defined inputs. They are useful for detecting arithmetic errors and for connecting coefficient patterns to counting subsets, as developed in Combinatorics.

Related topics

Combinations explains the selection counts in the coefficients. Power Series and Taylor and Maclaurin Series distinguish finite binomial identities from infinite local representations.