Binomial series again Watch

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Jorge
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#1
Report Thread starter 14 years ago
#1
Find the first 3 terms in the expansion of (1 + 3x^)^5, simplifying the coefficients. Given that the first 3 terms in the expansion of (a + bx) (1 + 3x)^5 are 3 + cx + 210 x^2, find the value of a, b and c.

I'm lost...

First part:1 + 15 x + 90 x

multiplying both

a + 15ax + 90ax^2 + bx + 15 bx^2 + 90 bx^3


Is that what I'm supposed to do?

Well a=1

the rest I can't work out.


Help....


Jorge
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kikzen
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#2
Report 14 years ago
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a + 15ax + 90ax^2 + bx + 15 bx^2 + 90 bx^3
=a + x(15a + b) + x^2(15b+90a)
a = 3
210= 15b+90a
b=-4

c = 15a+b
c= 41

sorry about that!
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Jorge
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#3
Report Thread starter 14 years ago
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Hi

Thanks, but you got me lost

where did you get the 105 from? Whre did you get the right hand side of the equation from?

Sorry but I'm dimmmmm.

Jorge
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Jorge
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#4
Report Thread starter 14 years ago
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Ok, I've managed to go ahead a bit

The two sides of the equation below must be equivalent

a + 15ax + 90ax^2 + bx + 15 bx^2 = 3 + cx + 210x^2
a + x(15a + b) + x^2(90a +15b) = 3 + cx + 210x^2

so

a= 3

90a + 15b = 270 + 15b = 210 .... b= -4

15a + b = c ... c= 45 - 4 = 41


In fact, while writing the message I ended up with the right answer.

I'll post it anyway, and thanks for your help.

Jorge
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