Question

How to derive this?\[a + a^2 + a^3\cdots a^n = {a(a^n -1) \over a - 1} \]

Answer 1

multiply by \(a\)

Answer 2

Everyone replying. Am I the only one who doesn't know this? lol

Answer 3

also it is wrong

Answer 4

Mathematica is not wrong!

Answer 5

denominator should be \(a-1\)

Answer 6

Sorry, yes.

Answer 7

I never make miskaes. lol

Answer 8

Let S represent the sum
\[S=a+a^2+a^3+...a^n\]
multiply this by a
\[aS= a^2+a^3+a^4+.....a^{n+1}\]
subtract S by aS
\[(1-a)S=a+0+0....-a^{n-1}\]
we have
\[(1-a)S=a-a^{n+1}\]
Can you find S now?

Answer 9

what @ash2326 said

Answer 10

Gee, how did you people become such geniuses?

Answer 11

And this is very similar to the geometric series sum proof.