Question

If the sum of coefficients in the expansion of $({\alpha }^2{x}^2+2\alpha x+1{)}^{51}$ is zero, then $\alpha$ is

• A. $1$
• B. $-1$
• C. $0$
• D. None of these

Answer 1

The correct option is B $-1$

As the sum of the coefficients of various terms in the expansion $({\alpha }^2{x}^2+2\alpha x+1{)}^{51}$ is zero, so putting $x=1$ in the expansion we have

${\alpha }^2+2\alpha +1=0$

$\Rightarrow (\alpha +1{)}^2=0$

$\Rightarrow \alpha =-1$

$\therefore$ Choice (b) is correct.