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I'm very confused. Would we have to expand the brackets?

If the first three terms of the expansion $(a-2x)^n$ are $1-16x+bx^2 - ...$ then find the values of $a$, $b$ and $n$?

Thank you so much!

Debbie Williams
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  • Yes, expanding and then comparing the suitable coefficients is the right way to go. – Stefan4024 Oct 08 '16 at 08:20
  • @Stefan4024 Yeah I expanded, but how do I, for example considering they are not numbers? Could you give me an example for example the second expansion? Because it's not $2-1$ for example, its $n-1$? – Debbie Williams Oct 08 '16 at 08:21

2 Answers2

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From the looks of it, it seems like the question asks for knowledge of the binomial theorem. It goes like this:

$ (x+y)^n = \sum_{k=0}^n {n \choose k} x^{n - k} y^k $

Now, you can use this to calculate for what $k$ and $n$ you get the right factor at $-16x$. Something we are able to deduce is that $a=1$. Seeing as the first term is $1$.

An interesting property of binomials is that - when raised to the $n$th power - the coefficients can be found in Pascal's triangle:

Pascal's triangle.

By using this fact, you'll easily be able to find the right value for $n$, and with it $b$.

Good luck!

Mitchell Faas
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As said, this is asking you to use the binomial expansion formulae. If you consider the lower order terms in $x$, you will show that $$(a-2x)^n=a^n-2 n a^{n-1}x+2n (n-1) a^{n-2} x^2+\cdots$$ which has to be equal to $1-16x+bx^2+\cdots$

Then, $????$

I am sure that you can take it from here.

Claude Leibovici
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