Question

In the expansion of


`(1 - (x)/(4) ) {}^(n)`
, the coefficient of

`x {}^(2)`
is

`(21)/(16)`
. Find the value of n.

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(-\fracx4\right)^2=(n!)/(2!(n-2)!)(x^2)/(16)`

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

`(n!)/(2!(n-2)!)=\frac{n(n-1)}2=21\implies n=7`