# How do you find the third term of (x+3)^12?

Jul 31, 2016

${3}^{r d}$ term of ${\left(x + 3\right)}^{12}$ is $594 {x}^{10}$

#### Explanation:

Any binomial expansion of the type ${\left(x + a\right)}^{n}$ has $n + 1$ terms starting from ${0}^{t h}$ term, and the ${n}^{t h}$ term is given by ""^nC_rx^(n-r)a^r, where ""^nC_r=(n!)/((n-r)!r!).

Hence ${3}^{r d}$ term of ${\left(x + 3\right)}^{12}$ will be with $n = 2$

""^12C_2x^(12-2)a^2

= (12!)/(10!2!)x^10a^2

= $\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{\left(10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\right) \left(2 \times 1\right)} {x}^{10} {a}^{2}$

= $\frac{12 \times 11 \times \cancel{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{\cancel{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \left(2 \times 1\right)} {x}^{10} {a}^{2}$

= $\frac{12 \times 11}{2 \times 1} {x}^{10} {a}^{2}$

= $6 \times 11 \times {x}^{10} {a}^{2}$

= $66 {x}^{10} {a}^{2}$

Hence ${3}^{r d}$ term of ${\left(x + 3\right)}^{12}$ will be $66 \times {x}^{10} \times {3}^{2} = 594 {x}^{10}$

Jul 31, 2016

$594 {x}^{10}$

#### Explanation:

${\left(a + b\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {a}^{n - k} {b}^{k}$

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

In our example $a = x$, $b = 3$ and $n = 12$

The first, second and third terms of ${\left(x + 3\right)}^{12}$ are:

$\left(\begin{matrix}12 \\ 0\end{matrix}\right) {x}^{12} = {x}^{12}$

$\left(\begin{matrix}12 \\ 1\end{matrix}\right) 3 {x}^{11} = \frac{12}{1} \cdot 3 {x}^{11} = 36 {x}^{11}$

$\left(\begin{matrix}12 \\ 2\end{matrix}\right) {3}^{2} {x}^{10} = \frac{12 \cdot 11}{2 \cdot 1} \cdot 9 {x}^{10} = 66 \cdot 9 {x}^{10} = 594 {x}^{10}$

If you were looking for more or later terms in the expansion, then it might be easier to pick them out from the appropriate row of Pascal's triangle. Some people call the first row of Pascal's triangle the "$0$th", which would make this the "$12$th" row. Others (like myself) call it the $13$th row. Let's just call it the row that starts $1 , 12$...

The terms on this row of Pascal's triangle are:

$\left(\begin{matrix}12 \\ 0\end{matrix}\right)$, $\left(\begin{matrix}12 \\ 1\end{matrix}\right)$, $\left(\begin{matrix}12 \\ 2\end{matrix}\right)$, ... $\left(\begin{matrix}12 \\ 12\end{matrix}\right)$.

So for example, the middle term of ${\left(x + 3\right)}^{12}$ is $924 \cdot {3}^{6} {x}^{6} = 673596 {x}^{6}$