Find the coefficient of x³ in the expansion of ≤ft[1/3(2 + x)]^{6}.

Correct Answer is : 160729160/729729160

Find the coefficient of x³ in the expansion of ≤ft[1/3(2 + x)]^{6}.

Correct Answer is : 160729160/729729160

Test Index

WAEC Further Mathematics 2016 Paper

Question : 15

Total: 40

x^{3}

in the expansion of

\left[\frac{1}{3}(2 + x)\right]^{6}

$\frac{135}{729}$
$\frac{149}{729}$
$\frac{152}{729}$
$\frac{160}{729}$

Validate

Solution:

$\left[\frac{1}{3}(2 + x)\right]^{6} = \left(\frac{2}{3} + \frac{x}{3}\right)^{6}$

The coefficient of $x^{3}$ is

${}^{6}C_{3}\left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right)^{3}x^{3} = \left(\frac{6!}{3!3!}\right)\left(\frac{8}{27}\right)\left(\frac{1}{27}\right)x^{3}$

= $\frac{160}{729}$

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