Coefficient of t^4 in expansion of {left( {frac{{1 - {t^6}}}{{1 - t}}} right)^3} is

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7.Binomial Theorem

hard

The coefficient of $t^4$ in the expansion of ${\left( {\frac{{1 - {t^6}}}{{1 - t}}} \right)^3}$ is

A

$12$

B

$15$

C

$10$

D

$14$

(JEE MAIN-2019)

Solution

$\left(1-t^{6}\right)^{3}(1-t)^{-3}$

$\left(1-t^{18}-3 t^{6}+3 t^{12}\right)(1-t)^{-3}$

$\Rightarrow$ coefficient of $t^{4}$ in $(1-t)^{-3}$ is

$^{3+1-1} C_{4}=^{6} C_{2}=15$

Standard 11

Mathematics

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