Question:
The coefficient of $t^{4}$ in the expansion of $\left(\frac{1-t^{6}}{1-t}\right)^{3}$
Correct Option: , 2
Solution:
Consider the expression
$\left(\frac{1-t^{6}}{1-t}\right)^{3}=\left(1-t^{6}\right)^{3}(1-t)^{-3}$
$\begin{aligned}=\left(1-3 t^{6}+3 t^{12}-t^{18}\right) &\left(1+3 t+\frac{3 \cdot 4}{2 !} t^{2}\right.\\ &\left.+\frac{3 \cdot 4 \cdot 5}{3 !} t^{3}+\frac{3 \cdot 4 \cdot 5 \cdot 6}{4 !} t^{4}+\mathrm{K} \infty\right) \end{aligned}$
Hence, the coefficient of $t^{4}=1 \cdot \frac{3 \cdot 4 \cdot 5 \cdot 6}{4 !}$
$=\frac{3 \times 4 \times 5 \times 6}{4 \times 3 \times 2 \times 1}$
$=15$