Middle terms, sums of coefficients and the term containing a given power of x
Answers are for parents and teachers.
1
Find the middle terms of (2x − 1/x)⁵.
n = 5, so T3 and T4. T3 = 5C2 (2x)³ (−1/x)² = 10 × 8x³ × x⁻² = 80x. T4 = 5C3 (2x)² (−1/x)³ = 10 × 4x² × (−x⁻³) = −40/x.
2
(a) Find the sum of the coefficients of (x − 3y)⁴. (b) The sum of the coefficients of (2x + 1)ⁿ is 2187. Find n.
(a) Put x = y = 1: (1 − 3)⁴ = 16. (b) 3ⁿ = 2187 = 3⁷, so n = 7.
3
The two middle terms of (4x + 3y)⁷ are equal. Find x/y.
They are equal when 4x = 3y, so x/y = 3/4.
4
Find the coefficient of x⁶ in (x² − 2/x)⁹. Is there a term in x⁷?
The power is 18 − 3r. For x⁶: r = 4 and the coefficient is 9C4 (−2)⁴ = 126 × 16 = 2016. For x⁷: r = 11/3, not whole, so there is no term in x⁷.
5
Find the coefficient of x⁵ in x²(3 + x)⁶.
It is the coefficient of x³ in (3 + x)⁶: 6C3 × 3³ = 20 × 27 = 540.