Middle terms, sums of coefficients and the term containing a given power of x
Week 2 · The middle term or terms of an expansion, mixed problems of Lesson 1, and finding the term that contains a given power of x · 3 lessons
1
The middle term, or the two middle terms, of an expansion
Pure Mathematics Sec 3 Term 1, Unit 1, Lessons 1–2
GoalFind the middle term, or the two middle terms, of an expansion.
New words
middle term · the term with as many terms before it as after itorder of a term · its place in the list T1, T2, T3, … (the order is r + 1)
Explanation
The expansion of (x + a)ⁿ has n + 1 terms. If n is even, n + 1 is odd, so there is exactly one middle term, number n/2 + 1 from the start. If n is odd, n + 1 is even, so there are two middle terms, number (n + 1)/2 and number (n + 3)/2. For n = 8 there are 9 terms and the middle one is T5 (r = 4); for n = 7 there are 8 terms and the middle ones are T4 and T5 (r = 3 and r = 4). Once you know the order, use the general term of Week 1. When you add or subtract two expansions, first count the terms that survive: (x + a)ⁿ + (x − a)ⁿ keeps only T1, T3, T5, …, and (x + a)ⁿ − (x − a)ⁿ keeps only T2, T4, …, so there are about half as many terms and the middle term is the middle one of that shorter list. The middle terms are where the numbers nCr are largest, because Pascal’s rows rise to the centre and then fall.
n
number of terms
middle term(s)
4
5
T3
5
6
T3 and T4
6
7
T4
7
8
T4 and T5
8
9
T5
9
10
T5 and T6
Worked examples
Find the middle term of (2x + 1/x)⁶. Here n = 6, so there is one middle term, T4 (r = 3): 6C3 (2x)³ (1/x)³ = 20 × 8x³ × x⁻³ = 160. It does not contain x.
Find the two middle terms of (x² + 1/x)⁷. Here n = 7, so they are T4 and T5. T4 = 7C3 (x²)⁴ (1/x)³ = 35x⁸ × x⁻³ = 35x⁵. T5 = 7C4 (x²)³ (1/x)⁴ = 35x⁶ × x⁻⁴ = 35x².
Find the middle term of (x + 1)⁶ − (x − 1)⁶. The difference is 2(T2 + T4 + T6): three terms, so the middle one is 2T4 = 2 × 6C3 x³ = 40x³. (Check: the whole difference is 12x⁵ + 40x³ + 12x.)
Class activity
Cross out from both ends. Write T1 to T9 for n = 8 and cross out the first and last, then the next pair, and so on until one remains. Do the same with T1 to T8 for n = 7 and see that two remain.
Practice
1
Find the middle term of (x + 1/x)¹⁰.
n = 10, so it is T6: 10C5 x⁵ x⁻⁵ = 252.
2
Give the order of the middle term(s) of an expansion with n = 11, and with n = 14.
n = 11: orders 6 and 7. n = 14: one middle term, order 14/2 + 1 = 8.