Answer:
[tex]3 + 3^2 + 3^3 ...+ 3^n = \frac{3(3^n - 1)}{2}[/tex]
Step-by-step explanation:
Given
[tex]3 + 3^2 + 3^3 ...+ 3^n[/tex]
Required
Show that the sum of the series is [tex]\frac{3(3^{n}-1)}{2}[/tex]
The above shows the sum of a geometric series and this will be calculated as shown below;
[tex]S_n = \frac{a(r^n - 1)}{r - 1}[/tex]
Where
a = First term;
[tex]a = 3[/tex]
r = common ratio
[tex]r = \frac{3^2}{3} = \frac{3^3}{3^2}[/tex]
[tex]r = 3[/tex]
Substitute 3 for r and 3 for a in the formula above;
[tex]S_n = \frac{a(r^n - 1)}{r - 1}[/tex] becomes
[tex]S_n = \frac{3(3^n - 1)}{3 - 1}[/tex]
[tex]S_n = \frac{3(3^n - 1)}{2}[/tex]
Hence;
[tex]3 + 3^2 + 3^3 ...+ 3^n = \frac{3(3^n - 1)}{2}[/tex]
