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# Not a question, just a useful formula

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

Feb 22, 2021

#1
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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
+112966
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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
+74
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i have learned about geometric progressions

pizzaisawesome972  Feb 24, 2021
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