Question
The coefficient of $$x^{13}$$ in the expansion of $${\left( {1 - x} \right)^5}{\left( {1 + x + {x^2} + {x^3}} \right)^4}$$ is
A.
$$4$$
B.
$$- 4$$
C.
$$0$$
D.
None of these
Answer :
$$4$$
Solution :
Expression $$ = {\left( {1 - x} \right)^5} \cdot {\left( {1 + x} \right)^4}{\left( {1 + {x^2}} \right)^4} = \left( {1 - x} \right){\left( {1 - {x^2}} \right)^4}{\left( {1 + {x^2}} \right)^4}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {1 - x} \right){\left( {1 - {x^4}} \right)^4}.$$
∴ the co-efficient of $${x^{13}} = - {\,^4}{C_3}{\left( { - 1} \right)^3} = 4.$$