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 :/