WAEC 2022 · Paper 1 · Q14

Find the coefficient of $x^2$ in the binomial expansion of $\left(x + \dfrac{2}{x^2}\right)^5$ .

Worked solution (try it first)

The general term is $^{5}C_r\, x^{5-r}\left(\dfrac{2}{x^2}\right)^r = {^{5}C_r}\, 2^r\, x^{5-3r}$ .
For $x^2$ you need $5 - 3r = 2$ , so $r = 1$ .
The coefficient is $^{5}C_1 \times 2^1 = 5 \times 2 = 10$ , option A.