☰ خطة العام · الرياضيات البحتة (بالإنجليزية)

Term 1 review: binomial expansions, complex numbers, and vectors, lines and planes

الأسبوع 15 · مراجعة الفصل الدراسي الأول: ذات الحدين والأعداد المركبة والمتجهات في الفراغ والمستقيمات والمستويات · 3 دروس
1

Review: binomial expansions, terms and coefficients

Pure Mathematics Sec 3 Term 1, Units 1–4, Review
GoalRevise the binomial theorem and choose the right tool for terms, coefficients, ratios and greatest terms.
New words
general term · T(r+1) = nCr xⁿ⁻ʳ aʳ, the term whose lower number is rfree term · the term with x⁰: the power of x in T(r+1) set to 0greatest term · found from the ratio T(r+1)/T(r) ≥ 1
Explanation

Weeks 1 to 3 give one theorem and several ways of using it. (x + a)ⁿ has n + 1 terms and its general term is T(r+1) = nCr xⁿ⁻ʳ aʳ, with r counted from 0. Read the question first: ‘term’ includes its power of x, ‘coefficient’ is the number in front, and ‘power’ is only the exponent. For a term counted from the end, convert it to a place counted from the start. For terms with a given power of x, write T(r+1) with every part of a in brackets, simplify the power of x, set it equal to k, solve for r, and accept r only if it is a whole number from 0 to n; the free term is the case k = 0. The middle term depends on whether n is even (one) or odd (two). Ratios use T(r+1)/T(r) = ((n − r + 1)/r) · (a/x); to find the greatest term, solve that ratio ≥ 1 for r, and when signs alternate compare the sizes first. Sums of coefficients come from x = 1. Keep the table as a checklist: before computing, decide which row your question belongs to.

tasktool
the (r + 1)th termT(r+1) = nCr xⁿ⁻ʳ aʳ
the kth term from the endterm number n + 2 − k from the start
the middle term(s)n even: T(n/2 + 1); n odd: T((n + 1)/2) and T((n + 3)/2)
the term with xᵏ, or free of xcollect the power of x, set it equal to k (or 0), solve for r: a whole number from 0 to n
the ratio of two termsT(r+1)/T(r) = ((n − r + 1)/r) · (a/x)
the greatest termsolve T(r+1)/T(r) ≥ 1 for r (compare sizes if the signs alternate)
the sum of the coefficientsput x = 1; the alternating sum: x = −1
Worked examples
In (2x − 3/x²)⁹, T(r+1) = 9Cr (2x)⁹⁻ʳ (−3/x²)ʳ, so the power of x is 9 − r − 2r = 9 − 3r. The free term needs 9 − 3r = 0, r = 3: 9C3 · 2⁶ · (−3)³ = 84 × 64 × (−27) = −145 152. The coefficient of x³ needs r = 2: 9C2 · 2⁷ · 9 = 36 × 128 × 9 = 41 472. For x², 9 − 3r = 2 gives r = 7/3: no such term.
The greatest term of (5 + 3x)⁷ at x = 1. T(r+1)/T(r) = ((8 − r)/r) × 3/5 ≥ 1 gives 3(8 − r) ≥ 5r, so r ≤ 3. At r = 3 the ratio is exactly 1, so T4 = T3 and both are greatest: 7C2 · 5⁵ · 3² = 21 × 3125 × 9 = 590 625, and 7C3 · 5⁴ · 3³ = 35 × 625 × 27 = 590 625.
Three consecutive coefficients of (1 + x)ⁿ are 120, 210, 252. With nC(r−1), nCr, nC(r+1): (n − r + 1)/r = 210/120 = 7/4 and (n − r)/(r + 1) = 252/210 = 6/5. The first gives 4n + 4 = 11r and the second 5n = 11r + 6. Subtracting gives n = 10, and then r = 4. Indeed 10C3 = 120, 10C4 = 210, 10C5 = 252.
Class activity

Build your own test. Choose a small binomial such as (x + 1/x)⁶. Write its seven terms, find the free term and the middle term, and check that the sum of the coefficients is 2⁶ = 64. Then give it to someone else to solve.

Practice
1
Find the third term from the end of (x + 2/x)¹¹.
2
Find the coefficient of x⁴ in (x² − 1/x)⁸.
3
Find the middle terms of (2x + y)⁷ and the relation between x and y that makes them equal.
4
In (1 + ax)⁸ the coefficient of x² equals the coefficient of x³. Find a.
5
Find the sum of the coefficients of (2x² + 1/x)⁶ and of (3x − 2)⁷.