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The question is: Find the binomial expansion of (1 + 2x)^3 up to and including the terms in x^3. I can do this.

I can't do this: State the range values of x for which these expansions are valid.

I thought you do:
|2x| < 1
|x| < 1/2
That's not what you do. What do you do? :frown:

That looks correct to me :s-smilie:

EDIT: oh wait, it would just be |x|<1 because its the type of binomial expansion that you found in C1/C2.
Reply 2
That's what I thought you do :confused:
But the answer just says "Expansion is finite and exact. Valid for all values of x." There's no modulus signs or equality signs.
Reply 4
(1+2x)31+6x+12x2+8x3(1 + 2x)^3 \equiv 1 + 6x + 12x^2 + 8x^3

It is to the power of a positive integer, so the expansion is exactly equal to it, so it works for all x. There's probably a better way to explain it, and your method would work if it was to the power of something else, but that's pretty much it.
Reply 5
Think about the expansion itself. As it is to the power of 3 this means that when you continue to use the binomial formula terms after x^3 will always be equal to zero (try it out). This is because if the power you are exploring is an integer then the expansion will work for all values of x. It is the same as rewriting (1+2x)^3 as (1+2x)(1+2x)(1+2x) which works for all x. If it still isn't clear tell me what you don't understand and i'll try to explain further.
Get Cape.Wear Cape.Fly.
But the answer just says "Expansion is finite and exact. Valid for all values of x." There's no modulus signs or equality signs.

Ah I see now. because the expansion is not to a fractional or negative power, It would stop after the 4th term. so there is no limitation of what values of x it is valid for.
IDGAF
Think about the expansion itself. As it is to the power of 3 this means that when you continue to use the binomial formula terms after x^3 will always be equal to zero (try it out). This is because if the power you are exploring is an integer then the expansion will work for all values of x. It is the same as rewriting (1+2x)^3 as (1+2x)(1+2x)(1+2x) which works for all x. If it still isn't clear tell me what you don't understand and i'll try to explain further.


Farhan.Hanif93
Ah I see now. because the expansion is not to a fractional or negative power, It would stop after the 4th term. so there is no limitation of what values of x it is valid for.


gcseeeman
(1+2x)31+6x+12x2+8x3(1 + 2x)^3 \equiv 1 + 6x + 12x^2 + 8x^3

It is to the power of a positive integer, so the expansion is exactly equal to it, so it works for all x. There's probably a better way to explain it, and your method would work if it was to the power of something else, but that's pretty much it.


So whenever it's an expansion to the power of a positive integer, it is valid for all values of x?
I think it has something to do with the power? All values of x apply if the power is a positive integer. You only use the modulus if the power is negative or a fraction. I think that's it anyway :smile:

If you do Edexcel C4 maths, check the summary section at the end of the chapter, it says it there (and I know this because I'm reading it now, haha :p:)

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