2013 InvB #42 a1=5, a2=4...

What could be so hard. It's a multiple-choice test for Pete's sake.

2013 InvB #42 a1=5, a2=4...

Postby turkeyhunt » Wed Apr 17, 2013 8:10 pm

If a1=5, a2=4 and a(n) = |(a(n-1) - a(n-2)|^(n-3) for n>=3, the a5=?

I get a(3) = |a(2)-a(1)|^0 = 1
a(4) = |a(3)-a(2)|^(4-3) = |1 - 4|^1 = 3
a(5) = |a(4) - a(3)|^(5-3) = |3 - 1|^2 = 4
a(5)=4

but the answer is 16. Can somebody explain what we're doing wrong?
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Re: 2013 InvB #42 a1=5, a2=4...

Postby rcoker7 » Wed Apr 17, 2013 8:35 pm

I read the original problem, and there's no absolute value sign. You should get a(3)=1. a(4) = -3. a(5) = 16
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Re: 2013 InvB #42 a1=5, a2=4...

Postby ZekeusAurelius » Wed Apr 17, 2013 8:36 pm

That's because there is no absolute value in the equation they give you, so instead of getting a4 = 3, you get a4 = -3, and so when you do (a4-a3)^2 instead of getting (3-1)^2 you get (-3-1)^2, which is 16
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