The coefficient of

Question:

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

  1. (1) 14

  2. (2) 15

  3. (3) 10

  4. (4) 12


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 $t^{4}=1 \cdot \frac{3 \cdot 4 \cdot 5 \cdot 6}{4 !}$

$=\frac{3 \times 4 \times 5 \times 6}{4 \times 3 \times 2 \times 1}$

$=15$

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