Question 912209
f(n)=5(1/2)^n


f(0)=5(1/2)^(0) ... plug in n = 0


f(0)=5(0.5)^(0)


f(0)=5(1)


f(0)=5



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f(n)=5(1/2)^n


f(1)=5(1/2)^(1) ... plug in n = 1


f(1)=5(0.5)^(1)


f(1)=5(0.5)


f(1)=2.5



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f(n)=5(1/2)^n


f(2)=5(1/2)^(2) ... plug in n = 2


f(2)=5(0.5)^(2)


f(2)=5(0.25)


f(2)=1.25



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f(n)=5(1/2)^n


f(3)=5(1/2)^(3) ... plug in n = 3


f(3)=5(0.5)^(3)


f(3)=5(0.125)


f(3)=0.625



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f(n)=5(1/2)^n


f(4)=5(1/2)^(4) ... plug in n = 4


f(4)=5(0.5)^(4)


f(4)=5(0.0625)


f(4)=0.3125



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The first five terms are: 5, 2.5, 1.25, 0.625, 0.3125