f(1) = 1
f(2) = 3·f(1) = 3·1 = 3
f(3) = 3·f(2) = 3·3 = 9
f(4) = 3·f(3) = 3·9 = 27
f(5) = 3·f(4) = 3·27 = 81
f(6) = ___?
Madi H.
asked 06/05/19f(1) = 1
f(2) = 3·f(1) = 3·1 = 3
f(3) = 3·f(2) = 3·3 = 9
f(4) = 3·f(3) = 3·9 = 27
f(5) = 3·f(4) = 3·27 = 81
f(6) = ___?
Ron G. answered 06/05/19
Multiple levels Math, Science, Writing
You might run into a scenario like this if you take a programming class and have to learn recursion. That's where the equation is defined by a previous (for instance) value of the same equation. You have to work your way back to a place where the equation has a known solution, and substitute that known solution, then work your way forward.
Start with n = 6 and work your way back.
f(6) = 3f(5)
f(5) = 3f(4)
f(4) = 3f(3)
f(3) = 3f(2)
f(2) = 3f(1) and since f(1) = 1, you use that knowledge to work your way forward...
f(2) = 3
f(3) = 9
f(4) = 27
f(5) = 81
f(6) = 243 ...and there you go.
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