Binomial Theorem And Its Simple Applications Question 125
Question: The coefficient of $ x^{10} $ in the expansion of $ {{(1+x^{2}-x^{3})}^{8}} $ is
Options:
A) 476
B) 496
C) 506
D) 528
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] We rewrite the given expression as $ {{[1+x^{2}(1-x)]}^{8}} $ and expand by using the binomial theorem. We have, $ {{[1+x^{2}(1-x)]}^{8}}{{=}^{8}}C_0{{+}^{8}}C_1x^{2}(1-x){{+}^{8}}C_2x^{4}{{(1-x)}^{2}} $
$ {{+}^{8}}C_3x^{6}{{(1-x)}^{3}}{{+}^{8}}C_4x^{8}{{(1-x)}^{4}} $
$ {{+}^{8}}C_5x^{10}{{(1-x)}^{5}}+… $
The two terms which contain $ x^{10} $ are $ ^{8}C_4 $
$ x^{8}{{(1-x)}^{8}} $ and $ ^{8}C_5x^{10}{{(1-x)}^{5}} $ .
Thus, the coefficient of $ x^{10} $ in the given expression is given by $ ^{8}C_4 $ [Coefficient of $ x^{2} $ in the expansion of $ {{(1-x)}^{4}}+{{[}^{8}}C_5 $
$ {{=}^{8}}C_4(6){{+}^{8}}C_5=\frac{8!}{4!4!}(6)+\frac{8!}{3!5!} $
$ =(70)(6)+56=476 $
BETA
