Binomial Theorem 1 Question 13
14.
The coefficient of $t^{4}$ in the expansion of $(\frac{1-t^{6}}{1-t})^3$ is
(a) 12
(b) 10
(c) 15
(d) 14
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Answer:
Correct Answer: 14. (c)
Solution:
- Clearly, $(\frac{1-t^{6}}{1-t})^3=\left(1-t^{6}\right)^{3}(1-t)^{-3}$
$\therefore$ Coefficient of $t^{4}$ in $\left(1-t^{6}\right)^{3}(1-t)^{-3}$
$=$ Coefficient of $t^{4}$ in $\left(1-t^{18}-3 t^{6}+3 t^{12}\right)(1-t)^{-3}$
$=$ Coefficient of $t^{4}$ in $(1-t)^{-3}$
$={ }^{3+4-1} C_{4}={ }^{6} C_{4}=15$
$ \left(\because \text { coefficient of } x^{r} \text { in }(1-x)^{-n}={ }^{n+r-1} C_{r}\right) $
