Answer:
D
Step-by-step explanation:
[tex]\sum\limits_{k=1}^38(\frac{1}{4})^{k-1}[/tex]
The symbol '[tex]\sum[/tex]' , read as sigma, is the symbol for summation. Since the expression below sigma is 'k= 1', while the number above sigma is 3, we are to find the sum of [tex]8(\frac{1}{4})^{k-1}[/tex] with each other from k =1 to k =3.
[tex]\boxed{\text{Multiple rule}: \sum\limits_{i=1}^nku_i=k\sum\limits_{i=1}^nu_i}[/tex]
[tex]\sum\limits_{k=1}^38(\frac{1}{4})^{k-1}[/tex]
= [tex]8\sum\limits_{k=1}^3(\frac{1}{4})^{k-1}[/tex] (Applying multiple rule)
= [tex]8[(\frac{1}{4})^{1-1}+(\frac{1}{4})^{2-1}+(\frac{1}{4})^{3-1}][/tex]
= [tex]8[(\frac{1}{4})^0+(\frac{1}{4})^1+(\frac{1}{4})^2][/tex]
= [tex]8(1+\frac{1}{4}+\frac{1}{16})[/tex]
= [tex]8(\frac{21}{16})[/tex]
= [tex]\bf{\frac{21}{2} }[/tex]
Thus, the answer is D.
