What is the formula of: $a^{0} + a^{1} + a^{2} + ... + a^{n-1} + a^{n}$?

Question

What is the formula of:

$$a^{0} + a^{1} + a^{2} + ... + a^{n-1} + a^{n}$$

Any ideas?

See the question Value of $\sum\limits_n x^n$, in particular this answer. — Américo Tavares, Apr 10 '13 at 16:06

score 3 · Answer 1 · answered Apr 10 '13 at 15:51

3

Hint: For $a=1$ this is simple. Otherwise, what happens when you multiply it by $a-1$?

answered Apr 10 '13 at 15:51

Cameron Buie

score 2 · Answer 2 · answered Apr 10 '13 at 15:50

2

See

it is called a gemoetric series and it is a standard result.

answered Apr 10 '13 at 15:50

Lost1

score 1 · Answer 3 · answered Apr 10 '13 at 15:51

1

If $a=1,a^{0} + a^{1} + a^{2} + ... + a^{n-1} + a^{n}=1+1+\cdots$ up to $(n+1)$ terms hence $=n+1$

Else let $S=a^{0} + a^{1} + a^{2} + ... + a^{n-1} + a^{n}$

So, $a\cdot S=a^{1} + a^{2} + a^{3} + ... + a^{n} + a^{n+1}$

So, $S(a-1)=a^{n+1}-1$

answered Apr 10 '13 at 15:51

lab bhattacharjee

Inceptio · Answer 4 · 2013-04-10T16:54:50.920

1

$$a^{n}-1=(a-1)(a^{n-1}+\cdots+1)\implies a^{n-1}+\cdots+1=\dfrac{a^{n}-1}{a-1}$$ where $a \neq 1$

edited Apr 10 '13 at 16:54

answered Apr 10 '13 at 15:59

Inceptio

4 Answers4