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If the constant term in the expansion of $\left(3 x^{3}-2 x^{2}+\frac{5}{x^{5}}\right)^{10}$ is $2^{k} \cdot l,$ where $l$ is an odd integer, then the value of $k$ is equal to :
Option: 1
$6$

Option: 2
$7$

Option: 3
$8$

Option: 4
$9$

Answers (1)

best_answer

$\mathrm{\left(3 x^{3}-2 x^{2}+\frac{5}{x^{5}}\right)^{10}} \\$

$\mathrm{= \frac{1}{x^{50}}\left(3 x^{8}-2 x^{7}+5\right)^{10}}$

Constant term in this expression

$\mathrm{=\text { coefficient of } x^{50} \text { in }\left(3 x^{8}-2 x^{7}+5\right)^{10}}$

General term in $\mathrm{\left(3 x^{8}-2 x^{7}+5\right)^{10}}$ is $\mathrm{\frac{10}{a ! b ! c !}\left(3 x^{8}\right)^{a}\left(-2 x^{7}\right)^{b}(5)^{c}}$

$\mathrm{=\frac{10}{a ! b ! c !} 3^{a} \cdot(-2)^{b} 5^{c} \cdot(x)^{8 a+7 b} }\\$

$\mathrm{\therefore \quad 8 a+7 b=50 \text { and } a+b+c=10 \text { and } a, b, c \in W }$

There is only one such set of values $\mathrm{a=1, b=6, c=3 }$

$\therefore$ Constant term is

$\mathrm{ \frac{10 !}{6 ! 3 !} 3^{1}(-2)^{6}(5)^{3}} \\$

$\mathrm{= \frac{10 \cdot 9 \cdot 8 \cdot 7}{5} \cdot 3^{1} \cdot 2^{6} \cdot 5^{3}} \\$

$\mathrm{= 2^{9} \cdot 5^{4} \cdot 21} \\$

$\mathrm{ k=9 }$

Hence the correct answer is option 4.

Posted by

Ajit Kumar Dubey

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