WAEC 2019 · Paper 2 · Q1

(a)
Given that $\binom nr = {}^nC_r$ , simplify $\binom{2x + 1}{3} - \binom{2x - 1}{3} - 2\binom x2$ .

Worked solution (try it first)

Write each term out: $\dbinom{2x + 1}{3} = \dfrac{(2x + 1)(2x)(2x - 1)}{6}$ and $\dbinom{2x - 1}{3} = \dfrac{(2x - 1)(2x - 2)(2x - 3)}{6}$ .
$2\dbinom x2 = 2 \times \dfrac{x(x - 1)}{2}$
$= x(x - 1)$ .
Multiply the whole expression by 6.
The first two terms share $(2x - 1)$ : $(2x - 1)[2x(2x + 1) - (2x - 2)(2x - 3)]$ .
The bracket: $4x^2 + 2x - (4x^2 - 10x + 6) = 12x - 6$
$= 6(2x - 1)$ , so the two terms give $6(2x - 1)^2 = 24x^2 - 24x + 6$ .
Take away $6x(x - 1) = 6x^2 - 6x$ : $18x^2 - 18x + 6$ .
Divide by 6: the expression is $3x^2 - 3x + 1$ .