[tex]a_1=4,\ a_2=16,\ a_3=64\\\\\text{a ratio}\ r=\dfrac{16}{4}=4\\\\\text{The formula of a sum of the geometric sequence}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\\text{substitute}\\\\S_8=4\cdot\dfrac{1-4^8}{1-4}=4\cdot\dfrac{1-65,536}{-3}=4\cdot\dfrac{-65,535}{-3}=4\cdot21,845=\boxed{87,380}[/tex]
Answer:
Sum is 87,380
Step-by-step explanation:
the sum of the geometric sequence 4, 16, 64,...... 8 terms
To find the sum of geometric sequence use formula
[tex]S_n = a_1 * \frac{1-r^n}{1-r}[/tex]
a_1 is the first term
r is the common ratio
To find out common ratio 'r', divide the second term by first term
16/4= 4
64/16= 4
r= 4
first term is also 4
plug in the values in the formula
[tex]S_n = 4 * \frac{1-4^8}{1-4}[/tex]
[tex]S_n = 4 * \frac{1-65536}{-3}=4*21845= 87380[/tex]
