The coefficient of t^4 in the expansion of {l | vedclass

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(B) The expression is given by $(1-t^6)^3 (1-t)^{-3}$.Expanding $(1-t^6)^3$ using the binomial theorem,we get $(1 - 3t^6 + 3t^{12} - t^{18})$.We need

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$15$

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(B) The expression is given by $(1-t^6)^3 (1-t)^{-3}$.Expanding $(1-t^6)^3$ using the binomial theorem,we get $(1 - 3t^6 + 3t^{12} - t^{18})$.We need the coefficient of $t^4$ in the product $(1 - 3t^6 + 3t^{12} - t^{18}) (1-t)^{-3}$.Since we only need the $t^4$ term,we only consider the constant term $1$ from the first bracket and multiply it by the coefficient of $t^4$ in the expansion of $(1-t)^{-3}$.The expansion of $(1-t)^{-n}$ is $\sum_{r=0}^{\infty} {^{n+r-1}C_r} t^r$.For $n=3$,the coefficient of $t^4$ is $^{3+4-1}C_4 = ^{6}C_4 = ^{6}C_2$.$^{6}C_2 = \frac{6 	imes 5}{2 	imes 1} = 15$.

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

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