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WAEC Further Mathematics 2018 Paper

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Question : 37

Total: 40

Find the constant term in the binomial expansion

\left(2x^{2} + \;\frac{1}{x}\right)^{9}

84
168
336
672

Solution:

Concept:

The constant term is the term in which the power of

x

is zero.

Explanation:

Write the general term of

\left(2x^2 + \frac{1}{x}\right)^9

as

T_{r+1} = \binom{9}{r}\left(2x^2\right)^{9-r}\left(x^{-1}\right)^r

Simplify the powers of

x

:

x^{2(9-r)}x^{-r} = x^{18-3r}

For the term to be constant, set the exponent equal to zero:

18 - 3r = 0 \Rightarrow r = 6

Now substitute

r = 6

:

T_7 = \binom{9}{6}\left(2x^2\right)^3\left(x^{-1}\right)^6

= \binom{9}{6} 2^3 = 84 \times 8 = 672

Answer:

D.

672

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