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C4 Binomial expansion

Taken from OCR C4 paper 2013 Q2
(9-16x)^3/2 expand first 3 terms. This gives 27-72x+32x^2
Next part I don't get....
State set of values for which the expansion is valid.
I can see why x<9/16 but I can't understand why x>-9/16 as stated in the mark scheme.
Reply 1
Avatar for james22
james22
16
What is the condition for (1+x)^r to have a valid expansion in general?

(here r is just any number e.g. 3/2).
Reply 2
Avatar for samsama
samsama
8
As a general rule, the expansion of a binomial with a negative or fractional index is only valid if |r|<1 (This can be proved using geometric progressions, but I forgot the proof.)
In this case |16x/9|<1, and therefore -9/16<x<9/16
Reply 3
Avatar for xlaser31
xlaser31
OP
6
Thank you for the replies, but I'm still unsure.
Can anybody else add light to this issue.
The examiners report simply states that many candidates avoided answering the last part (Valid set of values for x).
Reply 4
Avatar for TenOfThem
TenOfThem
18
Original post by xlaser31
Thank you for the replies, but I'm still unsure.
Can anybody else add light to this issue.
The examiners report simply states that many candidates avoided answering the last part (Valid set of values for x).


Do you understand the general condition

Do you realise that whilst 1+12+(12)2+(12)3+....1 + \frac{1}{2} + (\frac{1}{2})^2 + (\frac{1}{2})^3 + .... converges

1+3+32+33+....1 + 3 + 3^2 + 3^3 + .... does not



Do you realise that the fraction could be negative and still converge and that the 3 could be negative and still diverge



This gives us (1+r)n(1+r)^n converging if |r|<1, i.e. if -1<r<1
Reply 5
Avatar for TenOfThem
TenOfThem
18
Original post by xlaser31
Thank you for the replies, but I'm still unsure.
Can anybody else add light to this issue.
The examiners report simply states that many candidates avoided answering the last part (Valid set of values for x).


Do you understand the general condition

Do you realise that whilst 1+12+(12)2+(12)3+....1 + \frac{1}{2} + (\frac{1}{2})^2 + (\frac{1}{2})^3 + .... converges

1+3+32+33+....1 + 3 + 3^2 + 3^3 + .... does not



Do you realise that the fraction could be negative and still converge and that the 3 could be negative and still diverge



This gives us (1+r)n(1+r)^n converging if |r|<1, i.e. if -1<r<1 for all values of n

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