Binomial theorem: Formulas

Class 11

General term of a binomial expansion

T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}

What each letter means
$T_{r + 1}$	the term in position r + 1 of the expansion of (a + b)ⁿ
$n$	the power (a positive whole number)
$r$	one less than the term's position (0 for the first term)
$a$	the first term inside the bracket
$b$	the second term inside the bracket, with its sign

In the expansion of (a + b)ⁿ, the term in position r + 1 is ⁿCᵣ aⁿ⁻ʳ bʳ: the power of a falls from n while the power of b rises from 0.

Why it works

Multiplying out n brackets (a + b)(a + b)…(a + b), each term takes b from r of the brackets and a from the other n − r. There are ⁿCᵣ ways to choose which r brackets give b, so aⁿ⁻ʳbʳ appears ⁿCᵣ times.

When to use it

Finding one term of an expansion without writing it all out: a given term, the term with a given power of x, the middle term, or the term without x.

How to use it

Match the bracket to (a + b)ⁿ, keeping any minus sign inside b.
For the k-th term put r = k − 1.
Work out ⁿCᵣ aⁿ⁻ʳ bʳ.

To remember: The term's position is one ahead of r.

Other forms

Special case $T_{r + 1} =^{n} C_{r} x^{r}$ When the bracket is (1 + x)ⁿ, so a = 1 and b = x.
To remember itIn every term, the powers of a and b add up to n.

Worked examples

Story

Work out 102³ as (100 + 2)³, one term at a time. What is the second term?

$(1 3) \times 10 0^{2} \times 2 = 60000$
All four terms: 1000000 + 60000 + 1200 + 8, which is 1061208.

Answer: 60000

Picture

A cube of side a + b is cut into blocks, as (a + b)³ = a³ + 3a²b + 3ab² + b³ shows. With a = 3 and b = 1, what is the total volume of the a × a × b blocks (the second term)?

$(1 3) \times 3^{2} \times 1 = 27$
That is 3 blocks of 9 cubic units each.

Answer: 27

Direct

Find the 4th term of (x + 2)⁶.

The 4th term has r = 3: ⁶C₃ x³ 2³.
$(3 6) \times 2^{3} = 160$

Answer: 160x³

Reverse

Which term of (x + 3)⁸ contains x⁵? Find it.

Try it first, then show the working

x⁵ needs n − r = 5, so r = 3: it is the 4th term, ⁸C₃ x⁵ 3³.
$(3 8) \times 3^{3} = 1512$

Answer: The 4th term, 1512x⁵

Exam

Find the middle term of (x/3 + 9y)¹⁰.

Try it first, then show the working

n = 10 is even, so there is one middle term, the 6th, with r = 5.
It is ¹⁰C₅ (x/3)⁵ (9y)⁵.
$(5 10) \times \frac{9 ^{5}}{3 ^{5}} = 61236$

Answer: 61236x⁵y⁵

Common mistake: Taking r = k for the k-th term (the first term has r = 0), and dropping the sign of b: in (x − 2)ⁿ, b = −2.

Class 11

Sum of the binomial coefficients

S = 2^{n}

What each letter means
$S$	the sum of all the coefficients ⁿC₀ + ⁿC₁ + … + ⁿCₙ
$n$	the power, or the number of things to choose from

The coefficients of (a + b)ⁿ, which are the numbers in row n of Pascal's triangle, add up to 2ⁿ.

Why it works

Put a = b = 1 in (a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + … + ⁿCₙbⁿ. The left side is 2ⁿ and the right side is the sum of the coefficients. It also counts choices: each of n things is either taken or left, 2 ways each, so 2ⁿ ways to take any number of them.

When to use it

The number of subsets of a set, the number of ways to choose any number of things (2ⁿ − 1 if at least one), and checking an expansion.

How to use it

For the sum of all the ⁿCᵣ, write 2ⁿ.
If the empty choice is not allowed, take 1 away.

To remember: Each thing: in or out. Two ways, n times.

Other forms

Same formula, another form $S = (1 + 1)^{n}$ Put a = b = 1 in (a + b)ⁿ.
To remember itThe rows of Pascal's triangle add up to 1, 2, 4, 8, 16, …

Worked examples

Story

A pizza shop offers 7 toppings. How many different pizzas can be ordered, with any number of toppings (even none)?

Each topping is on or off.
$2^{7} = 128$

Answer: 128

Picture

Check the rule with row 4 of Pascal's triangle: 1, 4, 6, 4, 1.

$1 + 4 + 6 + 4 + 1 = 16$
$2^{4} = 16$

Answer: 16

Direct

Add ⁶C₀ + ⁶C₁ + ⁶C₂ + … + ⁶C₆.

$2^{6} = 64$

Answer: 64

Reverse

A set has 256 subsets. How many elements does it have?

Try it first, then show the working

$2^{8} = 256$

Answer: 8

Exam

In how many ways can one or more of 5 different books be chosen?

Try it first, then show the working

Each book is taken or left: 2⁵ ways, less the one way that takes no book.
$2^{5} - 1 = 31$

Answer: 31

Common mistake: Forgetting to take away the empty choice when "at least one" is asked.