Answer:
A
Step-by-step explanation:
Since i = 1 to 3, it is probably quicker to evaluate each sum, that is
4 × [tex]\frac{1}{2} ^{0}[/tex] + 4 × [tex]\frac{1}{2} ^{1}[/tex] + 4 × ([tex]\frac{1}{2}[/tex]))²
= (4 × 1) + (4 × [tex]\frac{1}{2}[/tex]) + (4 × [tex]\frac{1}{4}[/tex])
= 4 + 2 + 1
= 7 → A
