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If n is a positive integer, is the value of b-a at least

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VP
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If n is a positive integer, is the value of b-a at least [#permalink] New post 12 Jan 2005, 08:02
If n is a positive integer, is the value of b-a at least twice the value of 3^n - 2^n ?

1. a = 2^n+1 and b = 3^n+1

2. n =3
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 [#permalink] New post 12 Jan 2005, 08:48
Pick A.

1) b= 3^n+1, a = 2^n+1
b-a = 3^n-2^n
So, b-a is not greater than twice the value of 3^n-2^n. Sufficient.

2) n = 3
Insufficient
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 [#permalink] New post 12 Jan 2005, 10:50
Pick "A". I am assuming offcourse that you meant
b= 3^(n+1), a = 2^(n+1)

Statement 1 : for n = 1 we have (b-a) ==> 9 - 4 = 5
and 3^n - 2^n = 1....so ans is yes

n = 2 we have (b-a) ===> 27-8 = 19
and 3^n - 2^n = 1....so ans is yes

Statement 2 is insuff.


ALTERNATE APPROACH:

b-a = 3^(n+1) - 2^(n+1) = 3^n x 3 - 2^n x 2 = 2(3^n x 1.5 - 2^n)

For all +ve n interger (3^n x 1.5 - 2^n) will be > than 3^n-2^n....So statement 1 is suff.
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 [#permalink] New post 12 Jan 2005, 15:11
A . explanation as above.
  [#permalink] 12 Jan 2005, 15:11
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If n is a positive integer, is the value of b-a at least

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