Given recursive formula: f(n) =-6f(n-1), f(1)=1/2
f(1) = 1/2
f(2) = -6(1/2)
f(3) = = -6((1/2)(-6))=(1/2)(-6)2
f(4) = -6((1/2)(-6)2)=(1/2)(-6)3
f(5) = -6((1/2)(-6)3) = (1/2)(-6)4
...
f(n) = (1/2)(-6)n-1
Nila R.
asked 05/06/21f(1) = 1/2
f(n)= -6f(n-1)
Given recursive formula: f(n) =-6f(n-1), f(1)=1/2
f(1) = 1/2
f(2) = -6(1/2)
f(3) = = -6((1/2)(-6))=(1/2)(-6)2
f(4) = -6((1/2)(-6)2)=(1/2)(-6)3
f(5) = -6((1/2)(-6)3) = (1/2)(-6)4
...
f(n) = (1/2)(-6)n-1
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Nila R.
Thank you so much05/06/21