+74 \(3^0+3^1+3^2+3^3+3^4\cdots3^{(n-1)}+3^n=\frac{(3^{(n+1)}+1)}{2}\)
+1223 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.)
+118687 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 -
