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

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Binomial Theorem

The coefficient of

x^{18}

in the product

(1 + x) (1 - x)^{10} (1 + x + x^{2})^{9}

is:

Answer: (A)

Group the factors cleverly:

(1 + x) (1 - x) = 1 - x^{2}

and

(1 - x)^{9} (1 + x + x^{2})^{9} = [(1 - x) (1 + x + x^{2})]^{9} = (1 - x^{3})^{9}

, so the product is

(1 - x^{2}) (1 - x^{3})^{9}

. Expanding,

(1 - x^{3})^{9} = \sum_{k = 0}^{9} (k 9) (- 1)^{k} x^{3 k}

. To reach

x^{18}

we either take

x^{18}

from the second factor with the

1

, needing

3 k = 18

, i.e.

k = 6

with coefficient

(6 9) = 84

, or take

- x^{2}

times

x^{16}

, but

3 k = 16

has no integer solution. The coefficient of

x^{18}

is therefore

84

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option