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
3
Sums and differences of two expansions, and finding unknowns from given coefficients and ratios of terms
Pure Mathematics Sec 3 Term 1, Unit 1, Lesson 1
GoalAdd and subtract two expansions, recognise a disguised expansion, and find unknowns from coefficients.
New words
sum of two expansions · (x + a)ⁿ + (x − a)ⁿ = 2(T1 + T3 + T5 + …)difference of two expansions · (x + a)ⁿ − (x − a)ⁿ = 2(T2 + T4 + …)
Explanation
The expansions of (x + a)ⁿ and (x − a)ⁿ differ only in signs. The terms with even r (T1, T3, T5, …) are the same in both; the terms with odd r (T2, T4, …) change sign. So adding cancels the odd-r terms and doubles the others: (x + a)ⁿ + (x − a)ⁿ = 2(T1 + T3 + T5 + …). Subtracting does the opposite: (x + a)ⁿ − (x − a)ⁿ = 2(T2 + T4 + …), where the T’s are the terms of (x + a)ⁿ. The table shows it for n = 5. Sometimes a long sum is secretly one expansion. If it reads Aⁿ + nC1 Aⁿ⁻¹B + nC2 Aⁿ⁻²B² + … + Bⁿ, it equals (A + B)ⁿ; if the signs alternate it equals (A − B)ⁿ. Find A and B by looking at the powers that fall in one bracket and rise in the other. Finally, when a question gives a coefficient or a ratio of coefficients, turn each into an equation with nCr. Two pieces of information give two unknowns; dividing one equation by another often cancels what you do not need.
Term
(x + 1)⁵
(x − 1)⁵
Sum
Difference
T1
x⁵
x⁵
2x⁵
0
T2
5x⁴
−5x⁴
0
10x⁴
T3
10x³
10x³
20x³
0
T4
10x²
−10x²
0
20x²
T5
5x
5x
10x
0
T6
1
−1
0
2
Worked examples
Simplify (x + 1)⁵ + (x − 1)⁵ and (x + 1)⁵ − (x − 1)⁵. With T1 = x⁵, T3 = 10x³, T5 = 5x: the sum is 2(x⁵ + 10x³ + 5x) = 2x⁵ + 20x³ + 10x. With T2 = 5x⁴, T4 = 10x², T6 = 1: the difference is 2(5x⁴ + 10x² + 1) = 10x⁴ + 20x² + 2. Check with x = 2: 3⁵ + 1 = 244 and 64 + 160 + 20 = 244; for the difference 3⁵ − 1 = 242 and 10 × 16 + 20 × 4 + 2 = 242.
Find the 4th term, in ascending powers of x, of S = (2x + 1)⁵ + 5(2x + 1)⁴(1 − x) + 10(2x + 1)³(1 − x)² + 10(2x + 1)²(1 − x)³ + 5(2x + 1)(1 − x)⁴ + (1 − x)⁵. The coefficients 1, 5, 10, 10, 5, 1 are row 5, so S = [A + B]⁵ with A = 2x + 1 and B = 1 − x. A + B = x + 2, so S = (2 + x)⁵ in ascending powers of x. T4 = 5C3 · 2² · x³ = 10 × 4 x³ = 40x³.
In (1 + kx)ⁿ = 1 + 12x + 60x² + …, find n and k. The coefficient of x is nk = 12 and of x² is nC2 k² = 60. Divide the second by (nk)² = 144: [n(n − 1)/2 · k²]/(n²k²) = (n − 1)/(2n) = 60/144 = 5/12. So 12n − 12 = 10n, n = 6 and k = 12/6 = 2. Check: (1 + 2x)⁶ = 1 + 12x + 15 × 4x² + … = 1 + 12x + 60x² + …
Class activity
Check the rule. Expand (x + 2)³ and (x − 2)³ separately and then add and subtract them. Compare with 2(T1 + T3) and 2(T2 + T4).