# How do you expand (3+m/3)^5?

Aug 3, 2017

See a solution process below:

#### Explanation:

Solution for this problem is below:

We can use Pascal's triangle to solve this problem.

The triangle values for the exponent 5 are:

$\textcolor{red}{1} \textcolor{w h i t e}{\ldots \ldots \ldots} \textcolor{red}{5} \textcolor{w h i t e}{\ldots \ldots \ldots} \textcolor{red}{10} \textcolor{w h i t e}{\ldots \ldots \ldots} \textcolor{red}{10} \textcolor{w h i t e}{\ldots \ldots \ldots} \textcolor{red}{5} \textcolor{w h i t e}{\ldots \ldots \ldots} \textcolor{red}{1}$

Therefore ${\left(\textcolor{b l u e}{3} + \textcolor{g r e e n}{\frac{m}{3}}\right)}^{5}$ can be multiplied as:

$\textcolor{red}{1} \left({\textcolor{g r e e n}{\left(\frac{m}{3}\right)}}^{0} \cdot {\textcolor{b l u e}{3}}^{5}\right) + \textcolor{red}{5} \left({\textcolor{g r e e n}{\left(\frac{m}{3}\right)}}^{1} \cdot {\textcolor{b l u e}{3}}^{4}\right) + \textcolor{red}{10} \left({\textcolor{g r e e n}{\left(\frac{m}{3}\right)}}^{2} \cdot {\textcolor{b l u e}{3}}^{3}\right) + \textcolor{red}{10} \left({\textcolor{g r e e n}{\left(\frac{m}{3}\right)}}^{3} \cdot {\textcolor{b l u e}{3}}^{2}\right) + \textcolor{red}{5} \left({\textcolor{g r e e n}{\left(\frac{m}{3}\right)}}^{4} \cdot {\textcolor{b l u e}{3}}^{1}\right) + \textcolor{red}{1} \left({\textcolor{g r e e n}{\left(\frac{m}{3}\right)}}^{5} \cdot {\textcolor{b l u e}{3}}^{0}\right) \implies$

$\textcolor{red}{1} \left(\textcolor{g r e e n}{1} \cdot {\textcolor{b l u e}{3}}^{5}\right) + \textcolor{red}{5} \left(\textcolor{g r e e n}{m} \cdot {\textcolor{b l u e}{3}}^{3}\right) + \textcolor{red}{10} \left({\textcolor{g r e e n}{m}}^{2} \cdot {\textcolor{b l u e}{3}}^{1}\right) + \textcolor{red}{10} \left({\textcolor{g r e e n}{m}}^{3} \cdot {\textcolor{b l u e}{3}}^{-} 1\right) + \textcolor{red}{5} \left({\textcolor{g r e e n}{m}}^{4} \cdot {\textcolor{b l u e}{3}}^{-} 3\right) + \textcolor{red}{1} \left({\textcolor{g r e e n}{m}}^{5} \cdot {\textcolor{b l u e}{3}}^{-} 5\right) \implies$

$\textcolor{red}{1} \left(\textcolor{g r e e n}{1} \cdot \textcolor{b l u e}{243}\right) + \textcolor{red}{5} \left(\textcolor{g r e e n}{m} \cdot \textcolor{b l u e}{27}\right) + \textcolor{red}{10} \left({\textcolor{g r e e n}{m}}^{2} \cdot \textcolor{b l u e}{3}\right) + \textcolor{red}{10} \left({\textcolor{g r e e n}{m}}^{3} \cdot \textcolor{b l u e}{\frac{1}{3}}\right) + \textcolor{red}{5} \left({\textcolor{g r e e n}{m}}^{4} \cdot \textcolor{b l u e}{\frac{1}{27}}\right) + \textcolor{red}{1} \left({\textcolor{g r e e n}{m}}^{5} \cdot \textcolor{b l u e}{\frac{1}{243}}\right) \implies$

$243 + 135 m + 30 {m}^{2} + \frac{10}{3} {m}^{3} + \frac{5}{27} {m}^{4} + \frac{1}{243} {m}^{5}$

Or, in standard form:

$\frac{1}{243} {m}^{5} + \frac{5}{27} {m}^{4} + \frac{10}{3} {m}^{3} + 30 {m}^{2} + 135 m + 243$