Question

If the sum of the coefficients in the expansion of $(a^2{x}^2-6ax+11{)}^{10},$ where $a$ is constant, is $1024,$ then the value of $a$ is

• A. $5$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (D)

$3$

For sum of the coefficients, we substitute x=1
$(a^2-6a+11{)}^{10}=1024$
$(a^2-6a+11{)}^{10}=2^{10}$
$a^2-6a+11=2$
$a^2-6a+9=0$
$(a-3{)}^2=0$
$a=3$