Answer:
(a)[tex]S_6=1092[/tex]
(b)[tex]S_5=363[/tex]
Step-by-step explanation:
Question 1
Given the geometric series
3+15+75+375+...
[tex]a_1=3, r=15/3=75/15=5\\$Therefore using:\\S_n=\dfrac{a_1(1-r^n)}{1-r} \\S_6=\dfrac{3(1-3^6)}{1-3}=\dfrac{3(-728)}{-2}=\dfrac{-2184}{-2}\\\\S_6=1092[/tex]
Question 2
Given the series: [tex]\sum_{n=1}^5 3^n[/tex]
[tex]3^1=3;3^2=9;3^3=27\\$The series is 3+9+27+\cdots\\$Therefore: r=9/3=27/9=3[/tex]
[tex]a_1=3, r=3\\$Therefore using:\\S_n=\dfrac{a_1(1-r^n)}{1-r} \\S_5=\dfrac{3(1-3^5)}{1-3}=\dfrac{3(-242)}{-2}=\dfrac{-726}{-2}\\\\S_5=363[/tex]
