You have to count [tex]\sum\limits_{i=1}^{3}\left(4\cdot \left(\dfrac{1}{2}\right)^{i-1}\right) [/tex].
For i=1,
[tex]4\cdot \left(\dfrac{1}{2}\right)^{i-1}=4\cdot \left(\dfrac{1}{2}\right)^{1-1}=4\cdot \left(\dfrac{1}{2}\right)^{0}=4\cdot 1=4[/tex].
For i=2,
[tex]4\cdot \left(\dfrac{1}{2}\right)^{i-1}=4\cdot \left(\dfrac{1}{2}\right)^{2-1}=4\cdot \left(\dfrac{1}{2}\right)^{1}=4\cdot\dfrac{1}{2}=2[/tex].
For i=3,
[tex]4\cdot \left(\dfrac{1}{2}\right)^{i-1}=4\cdot \left(\dfrac{1}{2}\right)^{3-1}=4\cdot \left(\dfrac{1}{2}\right)^{2}=4\cdot \dfrac{1}{4}=1[/tex].
Then [tex]\sum\limits_{i=1}^{3}\left(4\cdot \left(\dfrac{1}{2}\right)^{i-1}\right) =4+2+1=7[/tex].
Answer: Correct choice is C.
