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C4 binomial expansion

Not really understanding last part of this Q

Any help is greatly appreciated image.jpg
You have to use the second part to do the third. What happens if you let x=1/3 in the 2nd expansion you got?
Reply 2
As said above. If x=1/3 then you have (from the second part of the question) (8+3(1/3))^{1/3} = .
Original post by Ano9901whichone
You have to use the second part to do the third. What happens if you let x=1/3 in the 2nd expansion you got?


Original post by B_9710
As said above. If x=1/3 then you have (from the second part of the question) (8+3(1/3))^{1/3} =
.

okay i see. thanks

could someone explain why the 'second part' of the question can be used? i.e. the 8+3x inside brackets

what is not valid about (1+8)^1/3 = 9^1/3 ?
Reply 4
Original post by hi-zen-berg
okay i see. thanks

could someone explain why the 'second part' of the question can be used? i.e. the 8+3x inside brackets

what is not valid about (1+8)^1/3 = 9^1/3 ?


I'm sure you've learnt that the binomial expansion of (1+x)n(1+x)^n for non-integer nn is only valid for x1|x| \leq 1. It should be written in your formula booklet as well, next to the expansion.

Sooo, can't really plug in x=8x=8.

On the other hand (8+3x)1/3(8 + 3x)^{1/3} is valid for 3x1|3x| \leq 1 so x=1/3x=1/3 is perfectly valid here.
Original post by Zacken
I'm sure you've learnt that the binomial expansion of (1+x)n(1+x)^n for non-integer nn is only valid for x1|x| \leq 1. It should be written in your formula booklet as well, next to the expansion.

Sooo, can't really plug in x=8x=8.

On the other hand (8+3x)1/3(8 + 3x)^{1/3} is valid for 3x1|3x| \leq 1 so x=1/3x=1/3 is perfectly valid here.


Ahh I see thanks; have learnt that but it was a bit abstract and didn't care aha
Reply 6
Original post by hi-zen-berg
Ahh I see thanks; have learnt that but it was a bit abstract and didn't care aha


Basically always think about what restrictions apply when it comes to plugging things into expansions! :smile:

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