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

pizzaisawesome972 Feb 22, 2021

#1**+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.)

CubeyThePenguin Feb 22, 2021

#2**+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 -

Melody Feb 22, 2021