We have to find the sum of the series has been given as.
[tex]\sum_{k=1}^{8}5(\frac{4}{3} )^{k-1}[/tex] -----(1)
If the series has been given as,
[tex]\sum_{n=1}^{k}a(r)^{n-1}[/tex] -------(2)
It's a geometric series with first term 'a' and common ratio 'r'.
Comparing both the expressions (1) and (2),
[tex]a=[/tex] 5
[tex]r=\frac{4}{3}[/tex]
Number of terms 'n' = 8
Sum of the k terms of this series is given by,
Sum = [tex]\frac{a(r^k-1)}{r-1}[/tex]
= [tex]\frac{5[(\frac{4}{3})^8-1)]}{\frac{4}{3}-1 }[/tex]
= [tex]\frac{5(9.98872-1)}{\frac{1}{3}}[/tex]
= [tex]\frac{44.9436}{\frac{1}{3} }[/tex]
= 134.83
Therefore, Option (B) is the correct option.
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