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The first four terms in the binomial expansion of (1+(x/3) 6 are 1+ax+bx 2+cx 3 . Find the values of the constants a, b and c, giving your answers in theor simplest form.
 Feb 2, 2015

Best Answer 

 #1
avatar+33615 
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The firstfour terms of (1 + m)6 are 1 + 6m + 6*5/2*m2 + 6*5*4/(3*2)*m3

 

So if m = x/3 

a = 6/3 = 2

b = 6*5/2*(1/32) = 30/18 = 5/3

c = 6*5*4/(3*2)*(1/33) = 20/27

.

 Feb 2, 2015

1+0 Answers

 #1
avatar+33615 
+5
Best Answer

The firstfour terms of (1 + m)6 are 1 + 6m + 6*5/2*m2 + 6*5*4/(3*2)*m3

 

So if m = x/3 

a = 6/3 = 2

b = 6*5/2*(1/32) = 30/18 = 5/3

c = 6*5*4/(3*2)*(1/33) = 20/27

.

Alan Feb 2, 2015

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