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WAEC Further Mathematics 2018 Paper

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Question : 13

Total: 40

Find the coefficient of

x^{3}

in the binomial expansion of

\left(x - \;\frac{3}{x^{2}}\right)^{9}

.

324
252
-252
-324

Solution:

Concept:

Use the general term of a binomial expansion to find the term containing

x^3

.

Explanation:

The expression is

\left(x - \frac{3}{x^2}\right)^9 = (x - 3x^{-2})^9

.

The general term is

{}^{9}C_{r}(x)^{9-r}(-3x^{-2})^{r}

.

Simplifying the powers of

x

:

x^{9-r} \cdot x^{-2r} = x^{9-3r}

.

We require the term containing

x^3

, so set

9 - 3r = 3

.

Thus,

3r = 6

, giving

r = 2

.

So the required term is

{}^{9}C_{2}(x)^{7}(-3x^{-2})^{2}

.

= {}^{9}C_{2} \cdot x^{7} \cdot 9x^{-4}

= \frac{9 \times 8}{2!} \cdot 9x^{3} = 36 \times 9x^{3} = 324x^{3}

.

Answer:

The coefficient of

x^3

is

324

, which is Option A.

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