The coefficient of x13 in the expansion of (1 - x)5(1 + x + x2 + x3)4 is

Question

The coefficient of x13 in the expansion of (1 - x)5(1 + x + x2 + x3)4 is (A) 24 (B) 12 (C) 6 (D) 4

Tardigrade Admin · Accepted Answer

Let, E=(1 - x)5((1 + x) (1 + x2))4 =(1 - x)(1 - x)4(1 + x)4(1 + x2)4 =(1 - x)(1 - x2)4(1 + x2)4 =(1 - x)(1 - x4)4 =(1 - x)(1 - 4 x4 + 6 x8 - 4 x12 + x16) Coefficient of x13 is (- 1)(- 4)=4

Q. The coefficient of $x^{13}$ in the expansion of $(1 - x)^{5} (1 + x + x^{2} + x^{3})^{4}$ is

$24$

$12$

$6$

$4$

Let, $E = (1 - x)^{5} ((1 + x) (1 + x^{2}))^{4}$
$= (1 - x) (1 - x)^{4} (1 + x)^{4} (1 + x^{2})^{4}$
$= (1 - x) (1 - x^{2})^{4} (1 + x^{2})^{4}$
$= (1 - x) (1 - x^{4})^{4}$
$= (1 - x) (1 - 4 x^{4} + 6 x^{8} - 4 x^{12} + x^{16})$
Coefficient of $x^{13}$ is $(- 1) (- 4) = 4$

Q. The coefficient of x13 in the expansion of (1−x)5(1+x+x2+x3)4 is

24

12

6

4

Let, E=(1−x)5((1+x)(1+x2))4 =(1−x)(1−x)4(1+x)4(1+x2)4 =(1−x)(1−x2)4(1+x2)4 =(1−x)(1−x4)4 =(1−x)(1−4x4+6x8−4x12+x16) Coefficient of x13 is (−1)(−4)=4

Q. The coefficient of $x^{13}$ in the expansion of $(1 - x)^{5} (1 + x + x^{2} + x^{3})^{4}$ is

$24$

$12$

$6$

$4$

Let, $E = (1 - x)^{5} ((1 + x) (1 + x^{2}))^{4}$
$= (1 - x) (1 - x)^{4} (1 + x)^{4} (1 + x^{2})^{4}$
$= (1 - x) (1 - x^{2})^{4} (1 + x^{2})^{4}$
$= (1 - x) (1 - x^{4})^{4}$
$= (1 - x) (1 - 4 x^{4} + 6 x^{8} - 4 x^{12} + x^{16})$
Coefficient of $x^{13}$ is $(- 1) (- 4) = 4$