The binomial theorem: expansion, general term and sums of expansions
Answers are for parents and teachers.
1
Expand (3x − 1)⁴.
81x⁴ − 4·27x³ + 6·9x² − 4·3x + 1 = 81x⁴ − 108x³ + 54x² − 12x + 1. (Check x = 1: 16 = 81 − 108 + 54 − 12 + 1.)
2
Find the 4th term of (3x − 1/x)⁷.
r = 3: 7C3 (3x)⁴ (−1/x)³ = 35 × 81x⁴ × (−x⁻³) = −2835x.
3
Find the 3rd term from the end of (x² + 1/x)⁹.
There are 10 terms, so it is number 9 + 2 − 3 = 8 from the start, r = 7: 9C7 (x²)² (1/x)⁷ = 36x⁴ × x⁻⁷ = 36x⁻³.
4
Simplify (2x + 1)⁴ − (2x − 1)⁴.
2(T2 + T4) = 2(4(2x)³ + 4(2x)) = 2(32x³ + 8x) = 64x³ + 16x. (Check x = 1: 3⁴ − 1⁴ = 80 and 64 + 16 = 80.)
5
The coefficient of x² in (1 + kx)⁸ is 700. Find the possible values of k and the coefficient of x in each case.
8C2 k² = 28k² = 700, so k² = 25 and k = 5 or k = −5. The coefficient of x is 8k = 40 or −40.