# How do you use the binomial series to expand  (x + 3)^12 ?

Dec 30, 2015

${\left(x + 3\right)}^{12}$

$= {\sum}_{k = 0}^{12} \left(\begin{matrix}12 \\ k\end{matrix}\right) {3}^{k} {x}^{12 - k}$

$= {x}^{12} + 36 {x}^{11} + 594 {x}^{10} + 5940 {x}^{9} + 40095 {x}^{8} + 192456 {x}^{7} + 673596 {x}^{6} + 1732104 {x}^{5} + 3247695 {x}^{4} + 4330260 {x}^{3} + 3897234 {x}^{2} + 2125764 x + 531441$

#### Explanation:

You can use Pascal's triangle to obtain the values of $\left(\begin{matrix}12 \\ 0\end{matrix}\right)$, $\left(\begin{matrix}12 \\ 1\end{matrix}\right)$,...,$\left(\begin{matrix}12 \\ 12\end{matrix}\right)$

Write out the row starting $1$, $12$,...

$1$, $12$, $66$, $220$, $495$, $792$, $924$, $792$, $495$, $220$, $66$, $12$, $1$

Write out ascending powers of $3$ from ${3}^{0}$ to ${3}^{12}$...

$1$, $3$, $9$, $27$, $81$, $243$, $729$, $2187$, $6561$, $19683$, $59049$, $177147$, $531441$

Multiply both sequences together to get:

$1$, $36$, $594$, $5940$, $40095$, $192456$, $673596$, $1732104$, $3247695$, $4330260$, $3897234$, $2125764$, $531441$

These are the coefficients we need since:

${\left(x + 3\right)}^{12} = {\sum}_{k = 0}^{12} \left(\begin{matrix}12 \\ k\end{matrix}\right) {3}^{k} {x}^{12 - k}$

So:

${\left(x + 3\right)}^{12}$

$= {x}^{12} + 36 {x}^{11} + 594 {x}^{10} + 5940 {x}^{9} + 40095 {x}^{8} + 192456 {x}^{7} + 673596 {x}^{6} + 1732104 {x}^{5} + 3247695 {x}^{4} + 4330260 {x}^{3} + 3897234 {x}^{2} + 2125764 x + 531441$