Question

What is the coefficient of the term in `x^9` in the expansion of `(3+x^3)^5` ?

Answer 1

`((5),(3)) 3^2 = 90`

Explanation:

The Binomial Theorem tells us that:

`(a+b)^n = sum_(k=0)^n ((n),(k)) a^(n-k) b^k`

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

So with `n=5`, `a=3` and `b=x^3` we find:

`(3+x^3)^5 = sum_(k=0)^5 ((5),(k)) 3^(5-k) (x^3)^k`

The term in `x^9` occurs when `k=3` giving us:

`((5),(color(blue)(3))) 3^(5-color(blue)(3)) (x^3)^color(blue)(3) = (5!)/(2!3!)*9x^9`

`color(white)(((5),(3)) 3^(5-3) (x^3)^3) = (5*4)/2*9x^9`

`color(white)(((5),(3)) 3^(5-3) (x^3)^3) = 90x^9`

So the coeffient of `x^9` is `90`

Instead of calculating `((5),(3)) = 10` directly, we can read it from Pascal's triangle, in the row starting `1, 5,...`