Option C:
[tex]\sum_{n=1}^{5} 3(0.2)^{n-1}=3.75[/tex]
Solution:
Given expression:
[tex]\sum_{n=1}^{5} 3(0.2)^{n-1}[/tex]
Summation means sum of the numbers.
Put n = 1 to 5 and sum all the numbers.
[tex]\sum_{n=1}^{5} 3(0.2)^{n-1}[/tex]
[tex]=3(0.2)^{1-1}+3(0.2)^{2-1}+3(0.2)^{3-1}+3(0.2)^{4-1}+3(0.2)^{5-1}[/tex]
[tex]=3(0.2)^{0}+3(0.2)^{1}+3(0.2)^{2}+3(0.2)^{3}+3(0.2)^{4}[/tex]
Using exponential rule: [tex]a^0=1[/tex], so that [tex]0.2^0=1[/tex]
[tex]=3(1)+0.6+0.12+0.024+0.0048[/tex]
[tex]=3.7488[/tex]
= 3.75
[tex]\sum_{n=1}^{5} 3(0.2)^{n-1}=3.75[/tex]
Option C is the correct answer.
Hence approximate value of [tex]\sum_{n=1}^{5} 3(0.2)^{n-1}[/tex] is 3.75.