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F(n)=1/2*4^n, n≥0I need helping finding the geometric sequence of this.

User Cunnel
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1 Answer

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Geometric sequence

an = a1 r ^(n-1)

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f(n)=1/2*4^n, n≥0

in this case a1= 1/2

r = 4

n= 0 (you just replace n)

f( 0 )=1/2*4^0

f(0) = 1/2 =1/ 2

n f(n)

0 f(0) = 1/ 2 f( 0 )=1/2*4^0

1 f(1) = 2 f( 0 )=1/2*4^2 = 4/2

2 f(2) = 8 f( 0 )=1/2*4^2 = 16/2

3 f(3) = 32 f( 0 )=1/2*4^2 =64/2

4 f(4) = 128 f( 0 )=1/2*4^2 = 256/2

User Rupert Angermeier
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