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  4. The sum of the coefficients in the expansion of (a2x2 - 6ax + 11)10, where a is constant, is 1024, then the value of a is

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

Binomial Theorem

Solution:

The sum of coefficients in the expansion
$(a^2x^2 - 6ax + 11)^{10}$ can be obtained by putting $x = 1$
$\Rightarrow (a^2(1)^2 - 6a(1) + (11))^{10} = 1024$
$\Rightarrow (a^2 - 6a + 11)^{10} = 2^{10}$
$\Rightarrow a^2 - 6a + 11 = 2$
$\Rightarrow a = 3$