JEE Main 2017 · Mathematics · Binomial Theorem · hard

The coefficient of x^8 in the expansion of (1+x+x^2+x^3^6 is

Single correct option

The coefficient of

x^8

in the expansion of

\left(1+x+x^2+x^3\right)^6

Show answer and solution

Answer: (B) $546$

Factorise

1+x+x^2+x^3=(1+x)(1+x^2)

, so the expression is

(1+x)^6(1+x^2)^6

. Choose

x^{2k}

from

(1+x^2)^6

with coefficient

\binom6k

, and then

x^{8-2k}

from

(1+x)^6

with coefficient

\binom6{8-2k}

. Non-zero terms occur for

k=1,2,3,4

\binom61\binom66=6

\binom62\binom64=225

\binom63\binom62=300

\binom64\binom60=15

. The sum is

6+225+300+15=546

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