The binomial expansion flashcards

Test yourself one card at a time: say the answer, flip, then mark it. Cards you know drop out; the rest come back until you know them. Your progress stays in this browser.

n! (n factorial)
n × (n − 1) × (n − 2) × … × 2 × 1 for a positive integer n; the number of ways to arrange n different objects in a line. 0! = 1.
ⁿCᵣ
n!/(r!(n − r)!): the number of ways to choose r objects from n when order does not matter. Also written as n over r in large round brackets.
Pascal's triangle
A triangle of numbers starting with 1; each row begins and ends with 1 and every other entry is the sum of the two above it. Row n lists ⁿC₀, ⁿC₁, …, ⁿCₙ.
Ascending powers of x
Terms written in the order constant, x, x², x³, …. 'The first four terms' stops at x³.
Coefficient (as opposed to term)
The number multiplying a power of x. The term includes the power: in −4320x³ the coefficient is −4320.
Binomial
An expression with two terms, such as a + bx. A binomial expansion multiplies out a power of it.
Binomial expansion of (a + b)ⁿ, n a positive integer
aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + … + ⁿCᵣaⁿ⁻ʳbʳ + … + bⁿ (n + 1 terms). (Yes)
General term of (a + bx)ⁿ
The term in xʳ is ⁿCᵣ aⁿ⁻ʳ (bx)ʳ. (Yes, as the general term of the expansion)
ⁿCᵣ in factorials
ⁿCᵣ = n!/(r!(n − r)!). (Yes, with the binomial expansion)
Pascal's rule
ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁. (No — learn it)
Symmetry of ⁿCᵣ
ⁿCᵣ = ⁿCₙ₋ᵣ: choosing r to take is the same as choosing n − r to leave. (No — derive it)
ⁿC₂ and ⁿC₃ as products in n
ⁿC₂ = n(n − 1)/2, ⁿC₃ = n(n − 1)(n − 2)/6. (No — derive it)
Tree of choices for (a + b)³
Choose a or b from each of 3 brackets: 8 paths. 1 path gives a³, 3 give a²b, 3 give ab², 1 gives b³, so the coefficients are 1, 3, 3, 1.
Rows 0 to 6 of Pascal's triangle
1 / 1 1 / 1 2 1 / 1 3 3 1 / 1 4 6 4 1 / 1 5 10 10 5 1 / 1 6 15 20 15 6 1.
Bar chart of the coefficients of (a − bx)ⁿ with a, b > 0
Bars alternate in sign (positive, negative, …) because the second term is negative.
Graph of (1 + bx)ⁿ against the polynomial from its first few terms
They are almost identical close to x = 0 and separate further away, which is why the approximation needs a small x.
Grid method for (p + qx)(a + bx)ⁿ
Rows p and qx, columns the binomial terms; the coefficient of xʳ is the sum of the cells whose powers add to r.
Why are the binomial coefficients ⁿCᵣ?
Expanding (a + b)ⁿ means picking a or b from each of n brackets. aⁿ⁻ʳbʳ appears once for each choice of which r brackets supply b: ⁿCᵣ ways.
Why must 0! = 1?
ⁿC₀ = 1 (one way to choose nothing) and the formula gives n!/(0! n!) = 1/0!, so 0! must be 1.
Why does a first-few-terms approximation work for small x?
Each power of a small x is much smaller than the one before, so the terms left out hardly change the total.
How can a product (p + qx)(a + bx)ⁿ produce an x² term?
p × (x² term) plus qx × (x term). Add every pair whose powers sum to 2.
Where do negative and fractional powers n belong?
P4: (1 + x)ⁿ becomes an infinite series, valid for |x| < 1. In P2, n is a positive integer and the expansion stops after n + 1 terms.
Link between the binomial expansion and the binomial distribution (S2)
P(X = r) = ⁿCᵣ pʳ(1 − p)ⁿ⁻ʳ is one term of the expansion of ((1 − p) + p)ⁿ; the terms add to 1ⁿ = 1.
Evaluate ⁸C₃ from the factorial definition.
8!/(3! 5!) = (8 × 7 × 6)/(3 × 2 × 1) = 56.
Expand (1 + 2x)⁵ fully.
1 + 10x + 40x² + 80x³ + 80x⁴ + 32x⁵ (row 5: 1 5 10 10 5 1, times powers of 2).
Coefficient of x² in (3 − x)⁶?
⁶C₂ × 3⁴ × (−1)² = 15 × 81 = 1215.
Coefficient of x³ in (2 + x/3)⁶, as an exact fraction?
⁶C₃ × 2³ × (1/3)³ = 20 × 8/27 = 160/27.
ⁿC₂ = 45. Find n.
n(n − 1)/2 = 45 → n² − n − 90 = 0 → (n − 10)(n + 9) = 0 → n = 10.
In (1 + kx)⁴ the coefficient of x is −4. Find k.
4k = −4, so k = −1.
Use the first three terms of (1 + x)⁶ to estimate 1.01⁶.
1 + 6x + 15x² with x = 0.01: 1 + 0.06 + 0.0015 = 1.0615 (calculator 1.06152…).
Coefficient of x² in (1 + x)(1 − 2x)⁵?
(1 − 2x)⁵ = 1 − 10x + 40x² + …; 1 × 40 + 1 × (−10) = 30.
(−3x)² inside an expansion
Bracket the whole second term: (−3x)² = 9x², not −3x² or −9x².
The term in x⁴ uses which r?
r = 4. It is the fifth term, because the expansion starts with x⁰.
Approximating a value such as 0.995¹⁰ from (1 − x/4)¹⁰
Solve 1 − x/4 = 0.995 first (x = 0.02); do not substitute 0.995 for x.
Two unknowns n and k from two coefficients
Write both equations, then divide one by the other to cancel factors; reject any n that is not a positive integer.
Quick check on a full expansion
Put x = 1: the coefficients must add to the value of the bracket, e.g. (2 + 1)⁵ = 243.

Practise this topicNotes Questions (28)Test builder