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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!

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

2 Answers 2

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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

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

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  • $\begingroup$ 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. $\endgroup$
    – Knowledge
    Dec 6, 2018 at 16:12
  • $\begingroup$ @DvirIhie But form $5$ to $8$ is like $0$ to $8$ minus $0$ to $4$, that's what I wrote. $\endgroup$
    – user
    Dec 6, 2018 at 16:16
  • $\begingroup$ oh my bad thank you !!! $\endgroup$
    – Knowledge
    Dec 6, 2018 at 16:18
  • $\begingroup$ @DvirIhie You are welcome! Bye $\endgroup$
    – user
    Dec 6, 2018 at 16:19
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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:

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

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