# Understanding the Geometric Series Formula: Solving for a Complex Variable

• N468989
In summary: in summary, this is a problem about calculating derivatives. you need to be familiar with the homogeneous property to solve it.
N468989
hi everyone,

i had a class where my teacher was talking about this but i don't know where this comes from, if anyone knows could you please give hint on the type of material this is. Like what type of mathbook i should be looking for.

the problem was:

Demonstrate the following expression:

$$$\displaystyle\sum_{n=0}^{N-1} \alpha^n = \left\{ \begin{array}{l l} N & \quad \mbox{, \alpha = 1}\\ \frac{1-\alpha^N}{1-\alpha} & \quad \mbox{, for any complex, \alpha \neq 1}\\ \end{array} \right.$$$

thankyou very much for any help.

tko_gx said:
Demonstrate the following expression:

$$$\displaystyle\sum_{n=0}^{N-1} \alpha^n = \left\{ \begin{array}{l l} N & \quad \mbox{, \alpha = 1}\\ \frac{1-\alpha^N}{1-\alpha} & \quad \mbox{, for any complex, \alpha \neq 1}\\ \end{array} \right.$$$

thankyou very much for any help.

Well, the case where $\alpha=1$ should be easy enough. To demonstrate the second case, simply multiply the summation by the fraction $\frac{1-\alpha}{1-\alpha}$. This will give you the denominator you need, and some subsequent simplifications to the numerator give you the final answer.

tko_gx said:
hi everyone,

i had a class where my teacher was talking about this but i don't know where this comes from, if anyone knows could you please give hint on the type of material this is. Like what type of mathbook i should be looking for.

the problem was:

Demonstrate the following expression:

$$$\displaystyle\sum_{n=0}^{N-1} \alpha^n = \left\{ \begin{array}{l l} N & \quad \mbox{, \alpha = 1}\\ \frac{1-\alpha^N}{1-\alpha} & \quad \mbox{, for any complex, \alpha \neq 1}\\ \end{array} \right.$$$

thankyou very much for any help.

Precalculus or calculus text.

CS

tko_gx said:
Demonstrate the following expression:

$$\sum_{n=0}^{N-1} \alpha^n = \frac{1-\alpha^N}{1-\alpha}$$

thank you very much for any help.

Well, the case where $\alpha=1$ should be easy enough. To demonstrate the second case, simply multiply the summation by the fraction $\frac{1-\alpha}{1-\alpha}$. This will give you the denominator you need, and some subsequent simplifications to the numerator give you the final answer.

in other words, ask yourself why the following is true:

$$(1-\alpha)\sum_{n=0}^{N-1} \alpha^n = \sum_{n=0}^{N-1} \alpha^n - \alpha\sum_{n=0}^{N-1} \alpha^n = 1 - \alpha^N$$

if you can figger out why that is true, you've solved your problem.

thanks for the help i found it in a calculus book. it´s mathematical induction.

rbj said:
in other words, ask yourself why the following is true:

$$(1-\alpha)\sum_{n=0}^{N-1} \alpha^n = \sum_{n=0}^{N-1} \alpha^n - \alpha\sum_{n=0}^{N-1} \alpha^n = 1 - \alpha^N$$

if you can figger out why that is true, you've solved your problem.

i can use the homogeneous property...and thus get the anwser

anyway thankyou all for helping out.

Last edited:

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