Question

The sum of the coefficients in the expansion of ${(a^2{x}^2-2ax+1)}^{51}$ vanishes, then the value of $\alpha$ is-

• A. $2$
• B. $-1$
• C. $1$
• D. $-2$

Answer 1

The correct option is C $1$

Put $x=1$ in ${\left(a^2{x}^2-2ax+1\right)}^{51}$,

${\left(a^2{\left(1\right)}^2-2a\left(1\right)+1\right)}^{51}={\left(a^2-2a+1\right)}^{51}$

Since the sum of the coefficients vanishes, then,

${\left(a^2-2a+1\right)}^{51}=0$

This is only possible if $a=1$, because,

$\left(1^2-2\left(1\right)+1\right)=1-2+1$

$=0$

Therefore, the value of $a$ is 1.