If the term independent of x in the expansion of
${\left( {{3 \over 2}{x^2} - {1 \over {3x}}} \right)^9}$ is k, then 18 k is equal to :
Solution
General term,
<br><br>$${T_{r + 1}} = {}^9{C_r}{\left[ {{3 \over 2}{x^2}} \right]^{9 - r}}{\left( { - {1 \over {3x}}} \right)^r}$$<br><br>$${T_{r + 1}} = {}^9{C_r}{\left[ {{3 \over 2}} \right]^{9 - r}}{\left( { - {1 \over 3}} \right)^r}{x^{18 - 3r}}$$<br><br>For independent of x <br><br>18 $-$ 3r = 0 $\Rightarrow$ r = 6<br><br>$\therefore$ $${T_7} = {}^9{C_6}{\left( {{3 \over 2}} \right)^3}{\left( { - {1 \over 3}} \right)^6} = {{21} \over {54}} = k$$<br><br>$\therefore$ $18k = {{21} \over {54}} \times 18 = 7$
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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