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jbmow
8.  You can graph this or use the summation function on the TI computer. 
mreijepatmat
∞
∑4(0.5)ⁿ⁻¹
n=1
For n = 1 → 4(0.5)¹⁻¹ = 4(0.5)⁰ = 4(1) = 4
For n = 2 → 4(0.5)²⁻¹ = 4(0.5)¹ = 4(0.5) = 4(0.5)
For n = 3 → 4(0.5)³⁻¹ = 4(0.5)² 
For n = 4 → 4(0.5)⁴⁻¹ = 4(0.5)³ 
We notice this is a geometric series with first term:
a=4 & r= 0.5
Since r<1, then the formula of the sum when n→∞ is:

Sum= a(1/1-r)
Sum = 4/(1-0.5)
Sum = 4/(1/2) = 4 x 2

 ∞
∑4(0.5)ⁿ⁻¹ = 8 (answer D)
n=1

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