Question

In the expansion of

Answer 1

By the binomial theorem, the $x^2$ term in the expansion of $\left(1-\frac x4\right)^n$ is

$\dbinom n21^{n-2}\left(-\dfracx4\right)^2=\dfrac{n!}{2!(n-2)!}\dfrac{x^2}{16}$

which suggests that the contribution of the binomial coefficient should make up the remaining factor of 21. That is,

$\dfrac{n!}{2!(n-2)!}=\dfrac{n(n-1)}2=21\implies n=7$