The term free of x, ratios of terms, and the greatest term
Answers are for parents and teachers.
1
Find the term free of x in (x³ + 1/x)¹².
The power is 3(12 − r) − r = 36 − 4r = 0, so r = 9 and the term is T10 = 12C9 = 220.
2
Show that (x² + 2/x⁴)¹⁰ has no term free of x, and say for which n there is one in (x² + 2/x⁴)ⁿ.
The power is 2(10 − r) − 4r = 20 − 6r = 0 gives r = 10/3, not whole, so there is no free term. In general 2n − 6r = 0, r = n/3: n must be a multiple of 3.
3
In (1 + 3x)⁸ find x so that T4 = T3.
T4 ÷ T3 with r = 3 is ((8 − 3 + 1)/3)(3x) = 6x = 1, so x = 1/6. (Check: T3 = 252x², T4 = 1512x³, and 1512x = 252 gives x = 1/6.)
4
The coefficients of three consecutive terms of (1 + x)ⁿ are 20, 15, 6. Find n.
(n − r + 1)/r = 15/20 = 3/4 gives 4n + 4 = 7r. (n − r)/(r + 1) = 6/15 = 2/5 gives 5n = 7r + 2. Subtract the first from the second: n − 4 = 2, so n = 6 (and r = 4). Check: 6C3 = 20, 6C4 = 15, 6C5 = 6.
5
Find the greatest term of (5 + 2x)¹⁰ when x = 1.
T(r+1)/T(r) = ((11 − r)/r)(2/5) ≥ 1 means 22 − 2r ≥ 5r, so r ≤ 22/7, about 3.14. The terms rise up to T4 = 10C3 · 5⁷ · 2³ = 120 × 78 125 × 8 = 75 000 000.