Proof of Exponential function $a^n$

Question

How can I prove that $$\sum_{n=N}^M a^n = \frac{a^M - a^{N+1}}{1-a}$$ for $a$ not equal to $1$ and $$\sum_{n=N}^M a^n = M-N+1$$ for $a = 1$

Second sum looks wrong. Should equal $M - N + 1$. – trystero Feb 03 '14 at 01:52 — trystero, Feb 03 '14 at 01:52

score 2 · Answer 1 · answered Feb 03 '14 at 01:42

2

Can you verify this ?

$$ (1-a)(1+a+a^2+\cdots + a^M)=1-a^{M+1} $$

answered Feb 03 '14 at 01:42

HK Lee

1

Yes I can, but I can't see where that gets me. – user1971854 Feb 03 '14 at 03:00
1

$(1-a)(1+a+\cdots + a^N) =1-a^{N+1}\ (M>N)$. So we consider the subtraction between two equations : $$ (1-a)(a^M + \cdots + a^{N+1} ) = a^{N+1} - a^{M+1}$$ – HK Lee Feb 03 '14 at 03:05

1 Answers1