+0  
 
+4
85
4
avatar+74 

\(3^0+3^1+3^2+3^3+3^4\cdots3^{(n-1)}+3^n=\frac{(3^{(n+1)}+1)}{2}\)

 
 Feb 22, 2021
 #1
avatar+944 
+2

This can be generalized:

 

\(a^0 + a^1 + \dots + a^n = \frac{1 - a^{n+1}}{1 - a}\)

 

(I derived this from the formula for the finite geometric series. If I've made a mistake, please let me know.)

 Feb 22, 2021
 #2
avatar+112966 
+2

Hi Pizza,

 

This is the sum of a GP  (Geometric progression) which maybe you have not learned about yet.

 

The first term, a=1       (3^0=1)

The common ratio r=3       (because you keep multiplying by 3)

there are n+1 terms here (see if you can understand why it is not n)

 

The formula for the sum of a GP is       \(S_n=\frac{a(r^{n}-1)}{r-1}\)

 

But in this case n needs to be replaced with n+1  because there are n+1 terms.

 

\(S_{n+1}=\frac{a(r^{n+1}-1)}{r-1}\\ S_{n+1}=\frac{1(3^{n+1}-1)}{3-1}\\ S_{n+1}=\frac{(3^{n+1}-1)}{2}\\\)

 

You seem to have a + where there should be a -

 Feb 22, 2021
 #3
avatar+74 
+1

i have learned about geometric progressions

pizzaisawesome972  Feb 24, 2021
 #4
avatar+112966 
+1

I'm pleased :)

Melody  Feb 24, 2021

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