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I always like to share with my students what each Math topic has to do with their everyday life, particularly their future.

Yesterday, I did **Binomial Expansion**. It's about finding the RIGHT partner (: It's about mastering the skills of finding the correct match based on a set of factors.

Just like this question, you see that there are 2 brackets. 1st bracket is good for now. 2nd bracket needs us to do some expansion work by using the FORMULAE (It's provided during **GCE O Level Exams** but some schools don't seem to provide it for their mid year, strange)

Oh yar, there is NO need to remember the **Binomial Expansion** Formulae!

Then one of the questions that often pop up is " When do we stop our expansion for [TEX](1+\frac {x}{3})^{12}[/TEX]?"

To answer this question, it depends on what type of partners you are looking for.

We are interested in the coefficients of & so from the first bracket :

To have ,

- 6 will pair up with from [TEX](1+\frac {x}{3})^{12}[/TEX]
- will pair up with the constant from [TEX](1+\frac {x}{3})^{12}[/TEX]
Similarly, to have ,

- 6 will pair up with from [TEX](1+\frac {x}{3})^{12}[/TEX]
- will pair up with the ,from [TEX](1+\frac {x}{3})^{12}[/TEX]
- will pair up with the constant from [TEX](1+\frac {x}{3})^{12}[/TEX]

Now, do you know where you stop the expansion of [TEX](1+\frac {x}{3})^{12}[/TEX]?

Stop at the term :) aka the 3rd term

Don Partho says

How does the 6 pair up with the x^2 in (1+x/3)^12? there is no x^2, only an x, surely if you times them together you won't get a coefficient for the x^2 value, but for the x.... Can you please explain :/