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<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>

Binomial expansion question

Just a quick question why is x valid for modulus of 1/2 and not 1/3?

Part b(ii)
Reply 1
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Svesh
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binomial.png
Original post by Svesh
...


(1+3x1+2x)2=(1+3x)2(1+2x)−2\left( \dfrac{1+3x}{1+2x} \right)^2 = (1+3x)^2(1+2x)^{-2}

Expansion of (1+3x)2(1+3x)^2 is valid for all xx but the expansion of (1+2x)−2(1+2x)^{-2} isn't.

Overall, the range of validity is where both of these expansions are valid. So if the first one is valid at all times, and the second one isn't, then clearly the problem reduces to finding out when the second expansion is valid.
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Svesh
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Original post by RDKGames
(1+3x1+2x)2=(1+3x)2(1+2x)−2\left( \dfrac{1+3x}{1+2x} \right)^2 = (1+3x)^2(1+2x)^{-2}

Expansion of (1+3x)2(1+3x)^2 is valid for all xx but the expansion of (1+2x)−2(1+2x)^{-2} isn't.

Overall, the range of validity is where both of these expansions are valid. So if the first one is valid at all times, and the second one isn't, then clearly the problem reduces to finding out when the second expansion is valid.

Ah ok that makes sense but why is all x valid for (1+3x)^2
Original post by Svesh
Ah ok that makes sense but why is all x valid for (1+3x)^2


Is it not clear? (1+3x)2=9x2+6x+1(1+3x)^2 = 9x^2 + 6x + 1 which obviously holds and converges for any xx value you put in. It doesn't "break down" for certain xx like the infinite expansion.
Reply 5
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Svesh
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Original post by RDKGames
Is it not clear? (1+3x)2=9x2+6x+1(1+3x)^2 = 9x^2 + 6x + 1 which obviously holds and converges for any xx value you put in. It doesn't "break down" for certain xx like the infinite expansion.

oh yh lol completely forgot. Just going back over c4 now and forgotten some knowledge. Thanks!!

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