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Tough Sequence Question

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by iggy614 » Mon Jul 13, 2009 12:49 pm
The infinite sequence a1, a2,...,an is defined such that an = (n+2) / n for all n ≥ 1. What is the product of the first 10 terms of the sequence?

(A) 45
(B) 66
(C) 90
(D) 121
(E) 132


i'll put the answer later on today
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by GMATQuantCoach » Mon Jul 13, 2009 4:13 pm
I will use _ to represent subscript.

a_n = (n+2)/n

a_n * a_(n+2) = (n+2)/n * (n+4)/(n+2) = (n+4)/n

a_n * a_(n+2) * a_(n+4) = (n+4)/n * (n+6)/(n+4) = (n+6)/n

.
.
.

a_n * a_(n+2) * a_(n+4) * ... * a_(n+8) = (n+10)/n

Then let n = 1

a_1 * a_3 * ...* a_7 * a_9 = (1 + 10)/1 = 11

Let n = 2

a_2 * a_4 * ... * a_8 * a_10 = (2 + 10)/2 = 6

Therefore

a_1 * a_2 * ... * a_9 * a_10 = 6 * 11 = 66

The answer is B.
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by tom4lax » Tue Jul 14, 2009 10:13 am
Think of it in terms of factorials (kind of).

Numerator starts at (1+2) or 3 and continues to (10+2) or 12.

Denominator starts at 1 and continues to 10.

Everything cancels out save for 11 and 12 in the numerator and 1 and 2 in the denominator. 132 / 2 = 66.

Answer is B.
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by iggy614 » Tue Jul 14, 2009 10:55 am
the correct answer is "B"

thanks for the help :)!

Iggy
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