The Binomial Theorem

Squaring a bracket

Multiply (a + b) by itself, every term by every term: ${\left(a+b\right)}^2=a^2+ab+$(a + b)² = a² + ab + ba $+b^2=a^2+2ab+b^2$+ b² = a² + 2ab + b². The coefficients are 1, 2 and 1. The middle one is 2 because ab turns up twice, once as a times b and once as b times a.

Drawn as a square of side a + b, the area splits into four pieces: a square $a^2$a², a square $b^2$b², and two rectangles, each a by b. The two rectangles are the 2ab.

Cubing it

To cube the bracket, multiply the square by one more bracket: ${\left(a+b\right)}^3=\left(a+b\right)\left(a^2+2ab+b^2\right)$(a + b)³ = (a + b)(a² + 2ab + b²).

Multiply every term by every term. a times the square gives $a^3+2a^{2}b+a{b}^2$a³ + 2a²b + ab². b times the square gives $a^{2}b+2a{b}^2+b^3$a²b + 2ab² + b³. Together that is $a^3+2a^{2}b+a{b}^2+a^{2}b+2a{b}^2+b^3$a³ + 2a²b + ab² + a²b + 2ab² + b³.

Now collect the like terms: $2a^{2}b+a^{2}b=3a^{2}b$2a²b + a²b = 3a²b and $a{b}^2+2a{b}^2=3a{b}^2$ab² + 2ab² = 3ab². So ${\left(a+b\right)}^3=a^3+3a^{2}b+3a{b}^2+b^3$(a + b)³ = a³ + 3a²b + 3ab² + b³, and the coefficients are 1, 3, 3 and 1.

A cube of side a + b, cut at a along every edge. Pull it apart and count the pieces: one cube $a^3$a³, three slabs $a^{2}b$a²b, three rods $a{b}^2$ab² and one small cube $b^3$b³. Those counts are the coefficients 1, 3, 3, 1. Choose n = 2 to see the square come apart into $a^2$a², two rectangles ab and $b^2$b².

Pascal’s triangle

Write the coefficients of each power in a row, starting with ${\left(a+b\right)}^0=1$(a + b)⁰ = 1 and ${\left(a+b\right)}^1=a+b$(a + b)¹ = a + b. Row 0 is 1. Row 1 is 1, 1. Row 2 is 1, 2, 1. Row 3 is 1, 3, 3, 1.

Each number is the sum of the two numbers above it, and each row starts and ends with 1. In row 3, 3 = 1 + 2 and 3 = 2 + 1. This arrangement is called Pascal’s triangle.

Why the two above add

Look again at the cube. The $a^{2}b$a²b term came from two places: $a^2$a² times b, with coefficient 1, and 2ab times a, with coefficient 2. So its coefficient is 1 + 2 = 3. Those are the two numbers above it in row 2.

The same happens in every row. Multiplying a row by one more bracket (a + b) sends each term two ways, once times a and once times b, and each new term collects one contribution from each of the two terms above it.

Every expansion from the triangle

So the triangle gives the coefficients without multiplying any brackets out. Row 4 is 1, 1 + 3, 3 + 3, 3 + 1, 1, which is 1, 4, 6, 4, 1. So ${\left(a+b\right)}^4=a^4+4a^{3}b+6a^2{b}^2+4a{b}^3+b^4$(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴. The power of a falls by one from term to term and the power of b rises by one, and in every term the two powers add up to 4.

Row 5 is 1, 5, 10, 10, 5, 1, and row 6 is 1, 6, 15, 20, 15, 6, 1. So ${\left(1+x\right)}^5=1+5x+10x^2+10x^3+5x^4+x^5$(1 + x)⁵ = 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵. Put x = 1 to check the row: 1 + 5 + 10 + 10 + 5 + 1 = 32, and $2^5=32$2⁵ = 32.

Rows 0 to 6 of Pascal’s triangle. Each entry is the one directly above it plus the one above and to the left: in row 6, 20 = 10 + 10 and 15 = 5 + 10.

Each entry counts choices

Multiplying out (a + b)(a + b) means choosing either a or b from each bracket and adding up every possible result. There are four ways to choose: a then a, a then b, b then a, and b then b. Two of them give ab, so ab has coefficient 2.

For ${\left(a+b\right)}^4$(a + b)⁴ there are four brackets. A term $a^2{b}^2$a²b² comes from choosing b from exactly two of the four brackets and a from the other two. The number of ways to choose 2 brackets out of 4 is 6: the first and second, first and third, first and fourth, second and third, second and fourth, or third and fourth. That is why the middle of row 4 is 6.

The choices for ${\left(a+b\right)}^2$(a + b)²: a or b from the first bracket, then a or b from the second. Two of the four paths, the colored one and the one below it, give ab, so the coefficient of ab is 2.

The general term

The number of ways to choose r things from n is written nCr. It can be worked out without the triangle: nCr = n! ÷ (r! × (n − r)!). For example, $4C2=4\times 3/2=6$4C2 = 4 × 3 / 2 = 6, and $5C2=5\times 4/2=10$5C2 = 5 × 4 / 2 = 10.

In ${\left(a+b\right)}^n$(a + b)ⁿ, a term with b to the power r comes from choosing which r of the n brackets give a b. The other n − r brackets give an a. So each term is $nCr{a}^{n-r}{b}^r$nCr aⁿ⁻ʳ bʳ, and the whole expansion is the sum of these terms for r = 0, 1, 2, up to n. This is the binomial theorem.

For the $x^3$x³ term of ${\left(1+x\right)}^5$(1 + x)⁵, choose 3 of the 5 brackets to give an x: $5C3=5\times 4\times 3/\left(3\times 2\times 1\right)=60/6=10$5C3 = 5 × 4 × 3 / (3 × 2 × 1) = 60 / 6 = 10, and every other bracket gives 1. So the term is $10x^3$10x³.

When the first term is not 1

The powers of each term in the bracket go into every coefficient. In ${\left(2+x\right)}^3$(2 + x)³ the row is 1, 3, 3, 1, and the powers of 2 fall from $2^3$2³ to $2^0$2⁰: ${\left(2+x\right)}^3=1\times 8+3\times 4\times x+3\times 2\times x^2+1\times x^3=8+12x+6x^2+x^3$(2 + x)³ = 1 × 8 + 3 × 4 × x + 3 × 2 × x² + 1 × x³ = 8 + 12x + 6x² + x³.

A minus sign belongs to its term. In ${\left(x-2\right)}^4$(x − 2)⁴ the second term is −2, and its odd powers are negative: ${\left(x-2\right)}^4=x^4+4x^3\left(-2\right)+6x^2{\left(-2\right)}^2+4x{\left(-2\right)}^3+{\left(-2\right)}^4=x^4-8x^3+24x^2-32x+16$(x − 2)⁴ = x⁴ + 4x³(−2) + 6x²(−2)² + 4x(−2)³ + (−2)⁴ = x⁴ − 8x³ + 24x² − 32x + 16. Check at x = 3: ${\left(3-2\right)}^4=1$(3 − 2)⁴ = 1, and 81 − 216 + 216 − 96 + 16 = 1.

One term can be found without the rest. For the $x^3$x³ term of ${\left(2x+3\right)}^5$(2x + 3)⁵, the 2x must be chosen from 3 brackets and the 3 from the other 2: $5C2\times {\left(2x\right)}^3\times 3^2=10\times 8x^3\times 9=720x^3$5C2 × (2x)³ × 3² = 10 × 8x³ × 9 = 720x³.

The usual mistakes

Using the row before. ${\left(a+b\right)}^3$(a + b)³ needs row 3, 1, 3, 3, 1. Row 2, 1, 2, 1, belongs to the square.

Taking 4C1 for the $a^2{b}^2$a²b² term of ${\left(a+b\right)}^4$(a + b)⁴. 4C1 = 4 counts the ways to choose one bracket, which gives the $a^{3}b$a³b term. The $a^2{b}^2$a²b² term needs two brackets chosen, 4C2 = 6.

Counting the choices in order. Choosing the first and then the third bracket is the same choice as the third and then the first, so 4 × 3 = 12 counts each choice twice; halve it to 6.

Multiplying instead of choosing. The coefficient of $x^3$x³ in ${\left(1+x\right)}^5$(1 + x)⁵ is 5C3 = 10, not 3 × 5 = 15.

Leaving out the powers of the first term. ${\left(2+x\right)}^3$(2 + x)³ is $8+12x+6x^2+x^3$8 + 12x + 6x² + x³, not $1+3x+3x^2+x^3$1 + 3x + 3x² + x³.

Two applications

In the first application below, six coin tosses are the six brackets of ${\left(1+x\right)}^6$(1 + x)⁶: an x for a head and a 1 for a tail. The coefficient of $x^2$x² counts the sequences with exactly two heads.

In the second, every length of a block is multiplied by 1 + x, so its volume is multiplied by ${\left(1+x\right)}^3=1+3x+3x^2+x^3$(1 + x)³ = 1 + 3x + 3x² + x³. When x is small, the first two terms give a close estimate.