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f(x) = 1-f(x-1) For positive integer x, if f(2)=12, then compute f(2006).
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Oh god lol
oops that x=2 one sec
\[f(2)=1-f(2-1)=1-f(1)=12=> f(1)=-11\]
\[f(3)=1-f(2)=1-12=-11\] \[f(4)=1-f(3)=1-(-11)=12\] \[f(5)=1-f(4)=1-12=-11\]
Thank you! But what about f(2006)? I'm confused on how to incorporate that..
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i think there may be a pattern
I see..
\[f(6)=1-f(5)=1-(-11)=12\] \[f(7)=1-f(6)=1-12=-11\]
see its either -11 or 12
all odds are -11 evens are 12
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right
Wow, thanks so much you guys! First time using this site and I love it already. (:
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