# What is the 4th term in the expansion of (y+4x^3)^4?

## (y+4x^3) ^4

Apr 12, 2018

$256 y {x}^{9}$

#### Explanation:

It's ${\left(a + b\right)}^{n} = \left[{\textcolor{w h i t e}{o}}^{n} {C}_{0} \cdot {a}^{n - 0} \cdot {b}^{0}\right] + \left[{\textcolor{w h i t e}{o}}^{n} {C}_{1} \cdot {a}^{n - 1} \cdot {b}^{1}\right] + \left[{\textcolor{w h i t e}{o}}^{n} {C}_{2} \cdot {a}^{n - 2} \cdot {b}^{2}\right] + \left[{\textcolor{w h i t e}{o}}^{n} {C}_{3} \cdot {a}^{n - 3} \cdot {b}^{3}\right] + \ldots \ldots \ldots$

First term $\textcolor{w h i t e}{w w}$ $\implies \left({\textcolor{w h i t e}{i}}^{4} {C}_{0}\right) \left({y}^{4}\right) {\left(4 {x}^{3}\right)}^{0}$
Second term $\textcolor{w h i t e}{i}$ $\implies \left({\textcolor{w h i t e}{i}}^{4} {C}_{1}\right) \left({y}^{3}\right) {\left(4 {x}^{3}\right)}^{1}$
Third term $\textcolor{w h i t e}{w u}$ $\implies \left({\textcolor{w h i t e}{i}}^{4} {C}_{2}\right) \left({y}^{2}\right) {\left(4 {x}^{3}\right)}^{2}$
Fourth term $\textcolor{w h i t e}{i r}$ $\implies \left({\textcolor{w h i t e}{i}}^{4} {C}_{3}\right) \left({y}^{1}\right) {\left(4 {x}^{3}\right)}^{3}$

Fourth term $= 4 \cdot y \cdot 64 {x}^{9}$ $\textcolor{w h i t e}{w w w w w w}$ $\left[\text{as } {\textcolor{w h i t e}{i}}^{4} {C}_{3} = 4\right]$

That is, $256 y {x}^{9}$

Apr 12, 2018

$\textcolor{b l u e}{256 y {x}^{9}}$

#### Explanation:

For binomial expansions of the form ${\left(y + x\right)}^{n}$, with $y$ descending, we have:

${\sum}_{r = 0}^{n} \left(\begin{matrix}n \\ r\end{matrix}\right) \left({y}^{n - r} {x}^{r}\right)$

Where:

((n),(r))=color(white)(0)^nC_(r)=(n!)/((r!(n-r)!)

We require the 4th term. The power of $y$ in the 4th term will be $1$:

So:

$4 - r = 1 \implies r = 3$

We use this value of $r$:

$\left(\begin{matrix}4 \\ 3\end{matrix}\right) \left({y}^{4 - 3} {\left(4 {x}^{3}\right)}^{3}\right)$

simplifying:

$\left(\begin{matrix}4 \\ 3\end{matrix}\right) \left(64 y {x}^{9}\right)$

((4),(3))=(4!)/(3!(4-3)!)=(4xx3xx2xx1)/(3xx2xx1xx1)=(4xxcancel(3)xxcancel(2)xxcancel(1))/(cancel(3)xxcancel(2)xxcancel(1)xx1)

$= \frac{4}{1} = 4$

$\therefore$

The 4th term is therefore:

$4 \cdot \left(64 y {x}^{9}\right)$

$\textcolor{b l u e}{256 y {x}^{9}}$