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