Aspire Faculty ID #15873 · Topic: JEE Main 2019 (12 April Morning Shift) · Just now
JEE Main 2019 (12 April Morning Shift)

The coefficient of $x^{18}$ in the product $(1+x)(1-x)^{10}(1+x+x^{2})^{9}$ is:

Solution

$(1+x)(1-x)^{10} = (1-x^2)(1-x)^9$ 
$(1-x)^9(1+x+x^2)^9 = [(1-x)(1+x+x^2)]^9 = (1-x^3)^9$ 
$\Rightarrow (1+x)(1-x)^{10}(1+x+x^2)^9 = (1-x^2)(1-x^3)^9$ 
 $(1-x^3)^9 = \sum_{k=0}^{9} \binom{9}{k}(-1)^k x^{3k}$ 

 For $x^{18}$: 
$3k = 18 \Rightarrow k=6$ 
Coefficient $= \binom{9}{6}(-1)^6 = \binom{9}{6} = 84$ 

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