Question

The coefficient of `x^2` in the expansion of `(3x + y)^n` is `324`. What is the value of `n`?

Answer 1

`n` has a value of `9`.

Explanation:

I will start this problem out using deduction.

If the coefficient of `x^2` is `324` then this must be the 3rd term (because we always include `t_0`).

The nth term in the binomial expansion is given by `t_(k + 1) = color(white)(two)_nC_k(x)^(n - k)y^k` in `(x + y)^n`. We know the value of `t_3`, and we know that `k =2`.

`324x^2 = color(white)(two)_nC_2(1)^(n - 2) xx (3x)^2`

`324x^2 = color(white)(two)_nC_2(1)^(n - 2) xx 9x^2`

`36 = color(white)(two)_nC_2(1)^(n - 2)`

However, since `1^n`, no matter the value of `n` always equals `1`, we can get rid of that part of the equation completely.

We are left with:

`36 = color(white)(two)_nC_2`

Expanding:

`36 = (n!)/((n - 2)!2!)`

Eliminate the factorial:

`36 = (n(n - 1)(n - 2)!)/((n - 2)!2!)`

`36 = (n(n - 1))/2`

`72 = n^2 - n`

`0 = n^2 - n - 72`

`0 = (n - 9)(n + 8)`

`n = 9 or -8`

A negative answer is not possible, so `n = 9`.

Hopefully this helps!

Answer 2

`n=-8` or `n = 9`

Explanation:

A polynomial `p(x) = a_0+a_1x+a_2x^2+cdots` has the property

`a_2=1/(2!)(d^2/(dx^2)p(x))_(x=0)` so

`d^2/(dx^2)p(x)=9n(n-1)(1+3x)^(n-2)` so

`a_2=1/(2!)9n(n-1)=324`

Solving for `n` we have

`n=-8` or `n = 9`