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

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Director
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If n is a positive integer, is the value of b a at least [#permalink]

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New post 21 Sep 2006, 12:33
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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

Kudos [?]: 129 [0], given: 0

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New post 21 Sep 2006, 12:44
For me (A)

The question is : b – a > 2*(3^n – 2^n) (1) ? with n >0

Statment 1:
a=2^n+1 and b=3^n+1

o n=0 <=> b-a = 3^1 - 2^1 = 1 and 2*(3^0 – 2^0) = 2*0 = 0, the equation (1) is verrified
o n=1 <=> b-a = 3^2 - 2^2 = 5 and 2*(3^1 – 2^1) = 2*1 = 2, the equation (1) is verrified
o n=2 <=> b-a = 3^3 - 2^3 = 19 and 2*(3^2 – 2^2) = 2*5 = 10, the equation (1) is verrified

and so on,

SUFF

Statment 2:
No information about b-a

INSUFF

Last edited by Fig on 21 Sep 2006, 14:10, edited 1 time in total.

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New post 21 Sep 2006, 14:09
Answer: A

S1:
b = 3^(n+1) = 3.3^n
a = 2^(n+1) = 2.2^n

b-a = 3.3^n - 2.2^n > 2.3^n - 2.2^n

or 3^n > 0

As n is +ve integer this is true for all n.
Sufficient.

S2: b-a > 38?
Not sufficient...

Answer: A

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Senior Manager
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New post 21 Sep 2006, 17:05
Going with A

(1)
Is it a=2^(n+1) and b=3^(n+1) ?
or a=2^n + 1 and b=3^n + 1

Assuming the latter
b=3^n + 1 (Eq 1)
a=2^n + 1 (Eq 2)

Eq1 - Eq2
b - a = (3^n + 1 ) - (2^n + 1 ) = 3^n - 2^n

(2) Insuff , no value for b and a.

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New post 21 Sep 2006, 17:09
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

q:

is b-a>= 2(3^n- 2^n)

from one

3^n+1 - 2^n+1 > 2(3^n- 2^n)
3^n+1 - 2^n+1 > 2*3^n - 2^n+1

ie 3^n+1>2*3^n....n is positive intiger least value is = 1

thus 3^2 > 2*3......suff

from two
clearlu insuff

My answer is A

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Director
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New post 21 Sep 2006, 23:53
The OA is ............... A :)

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  [#permalink] 21 Sep 2006, 23:53
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If n is a positive integer, is the value of b a at least

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