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7.Binomial Theorem
hard
$\left(\frac{1- t ^{6}}{1- t }\right)^{3}$ के प्रसार में $t ^{4}$ का गुणांक है
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
