Question

In the expansion of ${(1-x-x^2+x^3)}^6,$ the sum of the coefficients of $x$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

$f\left(x\right)={(1-x-x^2+x^3)}^6$
Sum of coefficients of $x$ is $f\left(1\right)$
$f\left(1\right)=(1-1-1+1{)}^6=0$

Note :
$1)$ Sum of the coefficients of even powers of $x$ is $\frac{f\left(1\right)+f(-1)}{2}$
$2)$ Sum of the coefficients of odd powers of $x$ is $\frac{f\left(1\right)-f(-1)}{2}$