Grade 10

The binomial theorem

The formula expanding (a+b)ⁿ into a sum of terms with binomial coefficients.

IntuitionIdea: what happens when you multiply (a + b) by itself many times?

Multiplying out $(a+b)^2=a^2+2ab+b^2$ and $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ by hand shows a pattern: the coefficients 1, 2, 1 and 1, 3, 3, 1 are exactly the rows of Pascal's triangle, and each row is built from the one above it by adding two neighbors.

Pascal's triangle of binomial coefficients

\binom{n}{k}

, colored by parity (odd in violet, even in amber): each entry is the sum of the two entries directly above it.

SchoolThe formal statement

Definition: Binomial coefficient

For integers $0\le k\le n$ , the binomial coefficient $\binom{n}{k}$ counts the number of ways to choose $k$ objects out of $n$ , and equals $\binom{n}{k}=\dfrac{n!}{k!(n-k)!}$ .

(a+b)^n=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^k

In the formula, $n$ is the (nonnegative integer) exponent, $k$ runs over every value from $0$ to $n$ , and each term $\binom{n}{k}a^{n-k}b^k$ picks a power of $a$ , a power of $b$ that together add up to $n$ , weighted by the binomial coefficient $\binom{n}{k}$ .

2^n=\sum_{k=0}^n\binom{n}{k}

The first few rows of the expansion
$n$	Expansion of $(a+b)^n$	Number of terms
$2$	$(a+b)^2=a^2+2ab+b^2$	$3$
$3$	$(a+b)^3=a^3+3a^2b+3ab^2+b^3$	$4$
$4$	$(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$	$5$
$5$	$(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$	$6$

UndergraduateProof by induction and the sum of coefficients

Theorem: The binomial theorem

For every nonnegative integer $n$ and all numbers $a$ , $b$ : $(a+b)^n=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^k$

Why is it true?

Expanding $(a+b)(a+b)\cdots(a+b)$ ( $n$ factors) means picking either $a$ or $b$ from every factor and multiplying; the coefficient of $a^{n-k}b^k$ is exactly the number of ways to pick $b$ from $k$ of the $n$ factors, which is $\binom{n}{k}$ .

Proof

We prove it by induction on $n$ . Base case $n=1$ : $(a+b)^1=a+b=\binom{1}{0}a+\binom{1}{1}b$ , which matches the formula since $\binom{1}{0}=\binom{1}{1}=1$ .

Inductive step: assume the formula holds for some $n$ , so $(a+b)^n=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^k$ . Multiply both sides by $(a+b)$ : $(a+b)^{n+1}=\sum_{k=0}^n\binom{n}{k}a^{n+1-k}b^k+\sum_{k=0}^n\binom{n}{k}a^{n-k}b^{k+1}$ .

Reindexing the second sum with $j=k+1$ and collecting the coefficient of $a^{n+1-j}b^j$ from both sums gives $\binom{n}{j}+\binom{n}{j-1}$ for each $j$ from $0$ to $n+1$ (with the convention $\binom{n}{-1}=\binom{n}{n+1}=0$ ).

By Pascal's rule $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$ , this sum equals $\binom{n+1}{j}$ , so $(a+b)^{n+1}=\sum_{j=0}^{n+1}\binom{n+1}{j}a^{n+1-j}b^j$ , completing the induction.

Theorem: Sum of the binomial coefficients

For every nonnegative integer $n$ : $2^n=\sum_{k=0}^n\binom{n}{k}$

Why is it true?

Setting $a$ and $b$ both equal to $1$ in the binomial theorem turns every term $a^{n-k}b^k$ into $1$ , so the whole sum collapses to a count of how many terms there are in total.

Proof

Substitute $a=1,\ b=1$ into the binomial theorem $(a+b)^n=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^k$ : the left side becomes $(1+1)^n=2^n$ , and the right side becomes $\sum_{k=0}^n\binom{n}{k}1^{n-k}1^k=\sum_{k=0}^n\binom{n}{k}$ , since $1$ raised to any power is $1$ .

Equating the two sides gives exactly $2^n=\sum_{k=0}^n\binom{n}{k}$ .

This identity also has a direct combinatorial meaning: $\binom{n}{k}$ counts the $k$ -element subsets of an $n$ -element set $S$ , so $\sum_{k=0}^n\binom{n}{k}$ counts every subset of $S$ of any size at all, i.e. the whole power set, which has exactly $2^n$ elements because each of the $n$ elements is independently either in or out of a subset.

UndergraduateReal-World Applications and Worked Examples

The binomial theorem is not just an algebra trick: it gives fast approximations in finance (compound growth), and it underlies counting arguments in computer science, engineering and probability wherever independent yes/no choices are combined.

Example: Approximating compound growth

A savings account earns 2% interest per year. Use the binomial theorem to approximate the growth factor after 10 years, $(1+0.02)^{10}$ , using only the first three terms of the expansion.

Solution

Write $1+0.02$ in place of $a+b$ with $a=1$ , $b=0.02$ , $n=10$ : by the binomial theorem the exact value is $(1+0.02)^{10}=\sum_{k=0}^{10}\binom{10}{k}(0.02)^k$ .

Since $0.02$ is small, later terms shrink fast, so keep only $k$ =0,1,2: $\binom{10}{0}+\binom{10}{1}(0.02)+\binom{10}{2}(0.02)^2$ .

Compute each term: $\binom{10}{0}=1$ , $\binom{10}{1}(0.02)=0.2$ , $\binom{10}{2}(0.02)^2=45\times0.0004=0.018$ .

Adding gives $1+0.2+0.018=1.218$ , so the account grows by about a factor of $1.218$ , i.e. about 21.8%, close to the exact value $1.2190\ldots$ — the binomial theorem turns a tedious 10-fold multiplication into three easy terms.

Example: Counting error patterns in a data packet

A network packet has $8$ independent bits, each of which may or may not be flipped by noise. An engineer designing an error-detecting code needs to know how many of the $2^8$ possible bit patterns have exactly $3$ flipped bits, and wants to check this against the total count using the sum-of-coefficients identity.

Solution

Model each bit as choosing $a$ (unflipped) or $b$ (flipped) in $(a+b)^8$ , so the number of patterns with exactly $k$ flipped bits is the coefficient of $a^{8-k}b^k$ , namely $\binom{8}{k}$ .

For exactly $3$ flipped bits, $k$ =3, so the count is $\binom{8}{3}=56$ .

To check the total, set $a$ = $b$ =1 as in the sum-of-coefficients theorem: $(1+1)^8=2^8$ , and $2^8=256$ counts every one of the $2^8$ possible bit patterns, split by how many bits are flipped.

So out of $256$ total patterns, $56$ have exactly $3$ errors — the same binomial coefficients used for algebra also drive the design of error-correcting codes.

What is $\binom{5}{2}$ ?

What is the coefficient of $x^2$ in the expansion of $(1+x)^4$ ?

What is the sum of all the coefficients in the expansion of $(a+b)^6$ ?

A network engineer wants to know how many 6-bit strings have exactly 4 bits set to 1. Using the binomial coefficient $\binom{6}{4}$ , what is this count?