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Ax(y) is an operation that adds 1 to y and then multiplies t

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by gmatter2012 » Tue Jul 10, 2012 2:48 am
Ax(y) is an operation that adds 1 to y and then multiplies the result by x. If x = 2/3, then Ax(Ax(Ax(Ax(Ax(x))))) is between

(A) 0 and ½
(B) ½ and 1
(C) 1 and 1½
(D) 1½ and 2
(E) 2 and 2½
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by Anurag@Gurome » Tue Jul 10, 2012 3:01 am
gmatter2012 wrote:Ax(y) is an operation that adds 1 to y and then multiplies the result by x. If x = 2/3, then Ax(Ax(Ax(Ax(Ax(x))))) is between
Algebraic Approach:

Ax(y) = (y + 1)*x
Ax(x) = (x + 1)*x = (x^2 + x)
Ax(Ax(x)) = Ax(x^2 + x) = (x^2 + x + 1)*x = (x^3 + x^2 + x)
Ax(Ax(Ax(x))) = Ax(x^3 + x^2 + x) = (x^3 + x^2 + x + 1)*x = (x^4 + x^3 + x^2 + x)
...
Ax(Ax(Ax(Ax(Ax(x))))) = (x^6 + x^5 + x^4 + x^3 + x^2 + x)

This is a geometric series with first term = x and common ratio = x.
Hence, the sum = x(1 - x^6)/(1 - x)

For x = 2/3, the sum = (2/3)*(1 - (2/3)^6)/(1 - 2/3) = (2/3)*((1 - (2/3)^6)/(1/3) = 2*(1 - (2/3)^6)

Now, (2/3)^6 will be a very small number with respect to 1, the sum will be smaller than 2 but very close to it.

The correct answer is D.
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by gmatter2012 » Tue Jul 10, 2012 3:22 am
Anurag@Gurome wrote:
gmatter2012 wrote:Ax(y) is an operation that adds 1 to y and then multiplies the result by x. If x = 2/3, then Ax(Ax(Ax(Ax(Ax(x))))) is between
Algebraic Approach:

Ax(y) = (y + 1)*x
Ax(x) = (x + 1)*x = (x^2 + x)
Ax(Ax(x)) = Ax(x^2 + x) = (x^2 + x + 1)*x = (x^3 + x^2 + x)
Ax(Ax(Ax(x))) = Ax(x^3 + x^2 + x) = (x^3 + x^2 + x + 1)*x = (x^4 + x^3 + x^2 + x)
...
Ax(Ax(Ax(Ax(Ax(x))))) = (x^6 + x^5 + x^4 + x^3 + x^2 + x)

This is a geometric series with first term = x and common ratio = x.
Hence, the sum = x(1 - x^6)/(1 - x)

For x = 2/3, the sum = (2/3)*(1 - (2/3)^6)/(1 - 2/3) = (2/3)*((1 - (2/3)^6)/(1/3) = 2*(1 - (2/3)^6)

Now, (2/3)^6 will be a very small number with respect to 1, the sum will be smaller than 2 but very close to it.

The correct answer is D.
Wonderful solution !!
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