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Posted by **Elisabeth** on Saturday, November 5, 2011 at 9:21pm.

a.f(n) = 5*(2)^(n-1) ; for n = 1,2,3,4,…

b.a1 = 1/3 ; an+1 = = 3*an

- Math -
**MathMate**, Saturday, November 5, 2011 at 9:35pma.

f(n)=5*2^(n-1)

f(1)=5*2^(1-1)=5*2^0=5

f(2)=5*2^(2-1)=5*2^1=10

f(3)=5*2^(3-1)=5*2^2=20

f(4)=5*2^(n-1)=5*2^3=40

...

b.

a1=1/3

a2=a1+1=3*a1=3*(1/3)=1

a3=a2+1=3*a2=3*1=3

...

get the idea?

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