Question

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

Answer 1

Answer:
Coefficient is 405.

Answer 2

Consider the binomial expansion of (1 + x)ⁿ, which is the binomial theorem.

$(1 + x)^{n} = \sum_{r = 0}^{n}\left(\begin{array}{ccc}n\\r\end{array}\right)(1)^{n - r}(x)^{r}$

Thus, we can say that the expansion of (3 + x)⁵ is:

$(3 + x)^{5} = \sum_{r = 0}^{5}\left(\begin{array}{ccc}5\\r\end{array}\right)(3)^{5 - r}(x)^{r}$

We can see that the only way to have an x-value in the expansion is when r = 1. Substituting r = 1 into the expansion, we get:

$\text{Coefficient of x: }\left(\begin{array}{ccc}5\\1\end{array}\right)(3)^{5 - 1}$
$= 5 \cdot 3^{4}$
$= 5 \cdot 81$
$= 405$

Thus, the coefficient of the x term is 405, based on our binomial theorem.