☰ خطة العام · الرياضيات البحتة (بالإنجليزية)

The binomial theorem: expansion, general term and sums of expansions

الأسبوع 1 · نظرية ذات الحدين: مثلث باسكال والتوافيق ومفكوك ذات الحدين والحد العام والحد المعدود من النهاية ومجموع مفكوكين وفرقهما · 3 دروس
1

Pascal’s triangle, combinations and the expansion of a binomial power

Pure Mathematics Sec 3 Term 1, Unit 1, Lesson 1
GoalExpand a binomial power with Pascal’s triangle and with combinations.
New words
binomial · a sum of two terms, such as x + acombination nCr · the number of ways to choose r objects out of nexpansion · the power multiplied out as a sum of terms
Explanation

A binomial is a sum of two terms, such as x + a. Multiply it by itself and a pattern appears: (x + a)² = x² + 2xa + a² and (x + a)³ = x³ + 3x²a + 3xa² + a³. In (x + a)ⁿ there are n + 1 terms; the powers of x fall from n to 0 while the powers of a rise from 0 to n, and in every term the two powers add up to n. The coefficients are the rows of Pascal’s triangle: each number is the sum of the two numbers above it. Instead of building the triangle row by row, we can name its entries. The entry in row n and place r (places are counted from 0) is the number of ways to choose r objects out of n, written nCr and read ‘n choose r’ (in Arabic التوافيق). Its value is n!/(r!(n − r)!), where n! = n × (n − 1) × … × 1 and 0! = 1. For quick work cancel: 8C3 = (8 × 7 × 6)/(3 × 2 × 1) = 56. Two useful facts: nC0 = nCn = 1 and nCr = nC(n − r). The numbers of row n add up to 2ⁿ, the number of all subsets of a set of n objects (each object is either in or out). The binomial theorem says (x + a)ⁿ = nC0 xⁿ + nC1 xⁿ⁻¹a + nC2 xⁿ⁻²a² + … + nCn aⁿ. For (x − a)ⁿ put −a in place of a, so the signs alternate: + − + − …

nnC0nC1nC2nC3nC4nC5
01
111
2121
31331
414641
515101051
Worked examples
Expand (x + 2)⁵. Row 5 of the triangle is 1, 5, 10, 10, 5, 1. Powers of 2 are 2, 4, 8, 16, 32. So (x + 2)⁵ = x⁵ + 5·2x⁴ + 10·4x³ + 10·8x² + 5·16x + 32 = x⁵ + 10x⁴ + 40x³ + 80x² + 80x + 32. Check with x = 1: 3⁵ = 243 and 1 + 10 + 40 + 80 + 80 + 32 = 243.
Expand (2x − 1)⁵. Replace a by −1 and use (2x) as the first term: (2x)⁵ − 5(2x)⁴ + 10(2x)³ − 10(2x)² + 5(2x) − 1 = 32x⁵ − 80x⁴ + 80x³ − 40x² + 10x − 1. The signs alternate because the powers of −1 alternate. Check with x = 1: 1⁵ = 1 and 32 − 80 + 80 − 40 + 10 − 1 = 1.
Combinations and the row sum. Choosing 3 players out of 10 for a football practice: 10C3 = (10 × 9 × 8)/(3 × 2 × 1) = 120 ways. By the symmetry nCr = nC(n − r), 10C7 is also 120. A team captain picks any number of 4 helpers, from none to all four: 4C0 + 4C1 + 4C2 + 4C3 + 4C4 = 1 + 4 + 6 + 4 + 1 = 16 = 2⁴ ways.
Class activity

Build row 6. Write row 5 (1, 5, 10, 10, 5, 1), then add neighbours to get 1, 6, 15, 20, 15, 6, 1. Add the row (it should be 64 = 2⁶) and check 6C2 = 15 with the formula.

Practice
1
Find 6C2, 6C4, 9C9 and 10C3.
2
Expand (a + 2b)³.
3
Expand (x − 2)⁴.
4
How many terms has the expansion of (3x + 5)¹²? How many subsets has a set of 7 objects?
5
Use x = 1 in the expansion of (1 + x)⁵ to show that 1 + 5 + 10 + 10 + 5 + 1 = 32.