Answer:
Step-by-step explanation:
hello : let : S this sum : S =4×(1/2)^(1-1)+4×(1/2)^(2-1)+4×(1/2)^(3-1)
S= 4×(1/2)^(0)+4×(1/2)^(1)+4×(1/2)^(2)
but : (1/2)^(0)= 1
(1/2)^(1) = 1/2
(1/2)^(2)= 1/4
so : S= 4×1 +4×(1/2)+4×(1/4) = 4+2+1 = 6
7
