Lesson 10 of 10 · 15 min

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Must-know facts

18 facts

1(a + b)ⁿ = Σ nCr aⁿ⁻ʳ bʳ, r from 0 to n, for a positive integer n.
2The expansion has n + 1 terms.
3Power of a falls from n to 0, power of b rises from 0 to n; they always add to n.
4Binomial coefficients nCr form Pascal's triangle; row n is nC0 … nCn.
5Pascal's rule nCr + nC(r − 1) = (n + 1)Cr builds each row from the one above.
6nCr = nC(n − r), so every row is symmetric.
7General term Tᵣ₊₁ = nCr aⁿ⁻ʳ bʳ; the (r + 1)th term has b to the power r.
8(x − y)ⁿ has alternating signs; the term with yʳ carries (−1)ʳ.
9(1 + x)ⁿ = nC0 + nC1 x + … + nCn xⁿ.
10nC0 + nC1 + … + nCn = 2ⁿ.
11nC0 − nC1 + nC2 − … + (−1)ⁿ nCn = 0.
12Sum of the coefficients of any expansion: put the variable equal to 1.
13Largest binomial coefficient of row n: the middle one, nC(n/2) for even n.
14Row 6: 1, 6, 15, 20, 15, 6, 1, total 64.
15(x + 2)⁶ = x⁶ + 12x⁵ + 60x⁴ + 160x³ + 240x² + 192x + 64.
1698⁵ = (100 − 2)⁵ = 9039207968.
176ⁿ − 5n leaves remainder 1 on division by 25.
18Pascal's triangle was known in India as Meru Prastara, given by Pingla.

Common traps

Where marks are lost

Expanding (2x + 3y)⁵ with 2x⁴ in place of (2x)⁴.

Keep each part in brackets: (2x)⁴ = 16x⁴. The number goes up to the power too.

Calling nCr aⁿ⁻ʳbʳ the rth term.

It is the (r + 1)th term, because r starts at 0. The 4th term has r = 3.

Giving 7C3 = 35 as the coefficient of x⁴ in (2x + 1)⁷.

The coefficient includes 2⁴ from (2x)⁴: 35 × 16 = 560. 35 is only the binomial coefficient.

Writing every sign as + in (x − 2y)⁵.

The b-part is −2y, so the term with yʳ carries (−1)ʳ: + − + − + −.

Saying (a + b)ⁿ has n terms.

There are n + 1 terms, from r = 0 to r = n.

Adding the coefficients of (x − 3)⁴ as 1 + 12 + 54 + 108 + 81.

Put x = 1: the sum is (1 − 3)⁴ = 16, because the signs alternate.

In (x² + 2/x)⁶, setting the power of x² equal to zero to find the term free of x.

Combine the powers first: x²⁽⁶⁻ʳ⁾ × x⁻ʳ = x¹²⁻³ʳ, then 12 − 3r = 0 gives r = 4.

Using only 1 + nx and calling the result exact.

1 + nx is the start of an estimate; the true value of (1 + x)ⁿ for x > 0 is larger because the other terms are positive.

Formulas

10 to know

Binomial theorem

(a + b)ⁿ = nC0 aⁿ + nC1 aⁿ⁻¹b + nC2 aⁿ⁻²b² + … + nCn bⁿ

n a positive integer; n + 1 terms.

Sigma form

(a + b)ⁿ = Σ nCk aⁿ⁻ᵏ bᵏ, k = 0 to n

a⁰ = b⁰ = 1.

Binomial coefficient

nCr = n!/(r!(n − r)!)

nC0 = nCn = 1.

General term

Tᵣ₊₁ = nCr aⁿ⁻ʳ bʳ

The (r + 1)th term, 0 ≤ r ≤ n.

Pascal's rule

nCr + nC(r − 1) = (n + 1)Cr

How each row of the triangle is built.

Symmetry

nCr = nC(n − r)

Terms equally far from the two ends have equal binomial coefficients.

Difference

(x − y)ⁿ = Σ (−1)ᵏ nCk xⁿ⁻ᵏ yᵏ

Signs alternate.

One plus x

(1 + x)ⁿ = nC0 + nC1 x + nC2 x² + … + nCn xⁿ

Coefficient of xʳ is nCr.

Row sum

nC0 + nC1 + … + nCn = 2ⁿ

Put x = 1 in (1 + x)ⁿ.

Alternating sum

nC0 − nC1 + nC2 − … + (−1)ⁿ nCn = 0

Put x = 1 in (1 − x)ⁿ.

Key terms

12 terms

Binomial: An expression with exactly two terms, such as a + b or 2x − 3.
Index: The power n to which the binomial is raised.
Expansion: The power written out as a sum of separate terms.
Binomial coefficient: The number nCr in front of aⁿ⁻ʳbʳ in the expansion of (a + b)ⁿ.
Pascal's triangle: The triangle of binomial coefficients in which each inside entry is the sum of the two above it.
Meru Prastara: The name of the same triangular arrangement in Pingla's work.
General term: Tᵣ₊₁ = nCr aⁿ⁻ʳbʳ, a formula for any single term of the expansion.
Coefficient: The whole number (with sign) multiplying a power of the variable, which can include powers of numbers inside the binomial.
Term independent of x: The term in which the power of x is zero.
Middle term: The term halfway along the expansion; one for even n, two for odd n.
Mathematical induction: Proof by showing a statement true for n = 1 and showing that truth for k forces truth for k + 1.
Sigma notation: Σ written with limits, meaning the sum of the terms for each value of the counter.