Egypt Secondary 3 Pure Mathematics, Week 3: Three consecutive coefficients, terms in a given ratio, and the greatest term

The term free of x, ratios of terms, and the greatest term

Week 3 · The term free of x, ratios of consecutive terms and of their coefficients, three consecutive coefficients and the greatest term · 3 lessons

Three consecutive coefficients, terms in a given ratio, and the greatest term

Pure Mathematics Sec 3 Term 1, Unit 1, Lessons 2–3

GoalUse consecutive coefficients and ratios to find unknowns, and find the greatest term.

New words

three consecutive coefficients · nC(r−1), nCr, nC(r+1); two ratios give n and rgreatest term · T(r+1) is greater than T(r) while the ratio T(r+1)/T(r) is at least 1

Explanation

Three consecutive coefficients of (1 + x)ⁿ are nC(r−1), nCr and nC(r+1). If they are given as numbers, form the two ratios: nCr / nC(r−1) = (n − r + 1)/r and nC(r+1) / nCr = (n − r)/(r + 1). Each is a known number, so you get two linear equations in n and r. When terms satisfy a relation, such as one being the average of its neighbours, write each term with nCr and simplify. For the greatest term of (a + b)ⁿ with numbers put in, use the ratio T(r+1)/T(r) = ((n − r + 1)/r)(b/a). The next term is bigger while this is at least 1 and smaller once it drops below 1. Solve the inequality for r: the terms rise up to the last whole r that satisfies it, and the term after that is the greatest. If the ratio equals exactly 1 for some r, two neighbouring terms are equal and both are the greatest. Take care when signs alternate, as in (2 − x)⁹ at x = 3: compare the sizes first, then decide which of the equal-size terms has the greater value, the positive one.

(4 + 3x)¹¹ with x = 1

r	T(r+1) ÷ T(r)	T(r+1) compared with T(r)
3	2.25	bigger
4	1.5	bigger
5	1.05	bigger
6	0.75	smaller
7	0.54	smaller

Worked examples

The coefficients of three consecutive terms of (1 + x)ⁿ are 28, 56, 70. Find n. Call them nC(r−1), nCr, nC(r+1). Then (n − r + 1)/r = 56/28 = 2, so n + 1 = 3r. And (n − r)/(r + 1) = 70/56 = 5/4, so 4n = 9r + 5. Put n = 3r − 1: 12r − 4 = 9r + 5, so r = 3 and n = 8. Check: 8C2 = 28, 8C3 = 56, 8C4 = 70.

The coefficients of T2, T3, T4 of (1 + x)ⁿ are in arithmetic progression (T3 is the average of the other two). Find n. The coefficients are nC1, nC2, nC3, so 2 · nC2 = nC1 + nC3. That is n(n − 1) = n + n(n − 1)(n − 2)/6. Divide by n and multiply by 6: 6(n − 1) = 6 + (n − 1)(n − 2), so n² − 9n + 14 = 0, and n = 2 or n = 7. For n = 2 there is no T4, so n = 7. Check: 7, 21, 35 and 2 × 21 = 7 + 35.

Find the greatest term of (4 + 3x)¹¹ when x = 1. T(r+1)/T(r) = ((12 − r)/r)(3/4) ≥ 1 means 36 − 3r ≥ 4r, so r ≤ 36/7, about 5.14. So T(r+1) ≥ T(r) for r up to 5, that is T6 ≥ T5, but T7 < T6. The greatest term is T6 = 11C5 · 4⁶ · 3⁵ = 462 × 4096 × 243 = 459 841 536.

Find the greatest value among the terms of (2 − x)⁹ when x = 3. The sizes follow T(r+1)/T(r) = ((10 − r)/r)(3/2). It equals 1 at r = 6 (30 − 3r = 2r gives r = 6), so T6 and T7 have the same size, and it is below 1 afterwards. But T6 has r = 5, an odd power of −3, so T6 = −489 888, while T7 = 9C6 · 2³ · 3⁶ = 84 × 8 × 729 = +489 888. The greatest value is T7 = 489 888.

Class activity

Rise and fall. Write the ratios (7 − r)/r × 2 for (1 + x)⁶ with x = 2, for r = 1 to 6, and mark where the terms stop rising. Which term is greatest?

Practice

Three neighbouring terms of (1 + x)ⁿ have the coefficients 36, 84, 126. Find n.

Find the greatest term of (1 + x)¹⁰ when x = 2.

The coefficients of T5, T6, T7 of (1 + x)ⁿ are in arithmetic progression. Find n.

Find the greatest term of (2 + x)⁸ when x = 3.

Show that (3 + 2x)⁴ has two equal greatest terms when x = 1, and find them.