# How to derive the formula of the sum of this finite series: $\sum_{n=M}^N a^n$

I would like to know how to arrive at the following result that my teacher wrote on the board. They did not explain how it was done. I am also not sure what this series is called. Is it perhaps a power series?

$$N>M :\sum_{n=M}^N a^n = \frac{a^M-a^{N+1}}{1-a},a\neq1$$ $$N>M: \sum_{n=M}^N a^n = N-M+1,a=1$$

I am quite lost since my teacher only wrote the above formulae without any derivation. Can someone help me understand why they are true? Thank you!

• Looks like an application of geometric series, i.e. $\sum_{i=0}^n a^i = \frac{1-a^{n+1}}{1-a}$. Dec 6, 2018 at 16:01
• geometric series Apr 5, 2023 at 0:02

The first one can be derived by the geometric series

$$\sum_{n=M}^N a^n =\sum_{n=0}^N a^n-\sum_{n=0}^{M-1} a^n$$

the second one is simply

$$\sum_{n=M}^N 1$$

• but you wrote from 0 to N i dont think its the same, sum of series for exm. from 5 till 8 its not like from 0 to 5 minus series from 0 to 7. Dec 6, 2018 at 16:12
• @DvirIhie But form $5$ to $8$ is like $0$ to $8$ minus $0$ to $4$, that's what I wrote.
– user
Dec 6, 2018 at 16:16
• oh my bad thank you !!! Dec 6, 2018 at 16:18
• @DvirIhie You are welcome! Bye
– user
Dec 6, 2018 at 16:19

Let us define

\begin{align} s_n = \sum_{n = M}^{N} a^n = a^M + a^{M+1} + \cdots + a^N. \end{align} \tag{1}

Hence \begin{align} a s_n = a^{M+1} + a^{M+2} + \cdots + a^{N+1} \end{align}. \tag{2}

By subtracting (2) from (1), we get \begin{align} s_n - a s_n = a^{M} - a^{N+1} \end{align}. Therefore \begin{align} s_n = \sum_{n = M}^{N} a^n = \frac{a^{M} - a^{N+1}}{1- a} & \text{, for } a\neq 1 \end{align}

The solution when $$a=1$$ is trivial:

\begin{align} \sum_{n = M}^{N} 1^n = \sum_{n = M}^{N} 1 = M-N+1 \end{align},