Question

How do you find the coefficient of `x` in the expansion of `(x+3)^5`?

Answer 1

Coefficient is 405.

Explanation:

We use the binomial theorem:

`(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k)y^k`

where `((n),(k)) = (n!)/(k!(n-k)!)`

For `n=5` and `y=3` we are looking for the coefficient of `x^1`. This means we need `n-k = 1 implies k = 4`.

`((5),(4))x^1*3^4 = (5!)/(4!1!)*81*x = 405x`