# Binomial series

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#1
find the first four terms in the expansion of (1-3x)^3/2. By substituting alpha suitable value of x, Find an approximation of 97^3/2.
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1 month ago
#2
Do you know how to write the binomial series of (2x-1)^2.
If you know, them the same rule applies for the first part of the question.
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#3
(Original post by abdullahAK)
Do you know how to write the binomial series of (2x-1)^2.
If you know, them the same rule applies for the first part of the question.
Yes I do know,thank you Am stuck on the second party
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1 month ago
#4
Equate (1-3x)^3/2 and 97^3/2.
Then, take root of 3/2 on both sides and you will find the value of x.
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1 month ago
#5
Yes I do know,thank you Am stuck on the second party
You generally need to choose a small value of x for reasons you must have covered on your course as well as a multiplier which matches the binomial with the expression you want to evaluate. By inspection, both should be reasonably clear.
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1 month ago
#6
Yes I do know,thank you Am stuck on the second party
So, you want 1-3x = 97, which would mean x = -32.

Well, that's not going to work as you need |3x|<1.

So we look for a factor that we can take out, which is a) easy to compute the power of 3/2 of, and b) which gives us |3x|<1.

An example we could try might be 4, so we'd get [4(97/4)]^(3/2) which becomes 4^(3/2) (97/4)^(3/2), and equals 8 (97/4)^(3/2)

So you'd want 1-3x = 97/4

Nope, still won't work.

What other factor could you try that would make 1-3x a lot smaller?

Edit: These sorts of question are better in study help, mathematics forum - I've requested the mods to move it.
Last edited by ghostwalker; 1 month ago
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