anonymous
  • anonymous
The first three terms in the expansion of(2+ax) n, in ascending powers of x, are 32−40x+bx2. Find the values of the constants n, a and b.
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
expanding (2+ax)^n we have 2^n + n*a*x +n*(n-1)/2 *a^2x^2 ..... thus we have 2^n =32 but we know 32 =2^5 hence n=5
anonymous
  • anonymous
again n*a=-40 as we have n=5 hence we have a=-8
anonymous
  • anonymous
thus b=n*(n-1)/2 *a^2 =5*4/2 *(-8)^2=640

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anonymous
  • anonymous
@matricked I get the first step for the second binomial term isn't it n*2^(n-1)ax=-40x?
anonymous
  • anonymous
The answers say a = -1/2 and b = 20 . I understand how a = -1/2 but I don't for b
campbell_st
  • campbell_st
well n = 5 since 2^5 = 32 the binomial expansion is \[^5Cn(2)^{5 - n}(bx)^n\] where n = 0, 1, 2, 3, 4, 5
anonymous
  • anonymous
So to find the 3rd term do I do 5C2(2)^3(ax)^2?
campbell_st
  • campbell_st
oops its \[^5C_{1} (2)^4 (ax)^1 = -40x\] solve for x
anonymous
  • anonymous
Yes I understand that
campbell_st
  • campbell_st
which is \[5 \times 16 \times ax = -40x\]
anonymous
  • anonymous
So a = -0.5?
campbell_st
  • campbell_st
so the 3rd term is \[^5C_{2} (2)^3 (-\frac{1}{2}x)^2 = bx^2\]
campbell_st
  • campbell_st
yep a is definitely -1/2 so the last part is \[10 \times 8 \times (-\frac{1}{2})^2 \times x^2 = bx^2\]
anonymous
  • anonymous
And I presume b = 20. Thanks so much for your help!
campbell_st
  • campbell_st
well yes it does show b = 20... glad to help. \[\frac{80}{4}x^2 = bx^2\]

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