Proof
(related to Proposition: Sum of Geometric Progression)
- Let \(n\in\mathbb N\) be a natural number and \(a, x \in F\) be any two elements of a given field \((F, +, \cdot)\) with \(x\neq 1\).
- Set the sum \(S_n\) geometric progression \(a, ax, ax^2,\ldots ax^n\)
\[S_n:=\sum_{0\le k\le n} ax^k\]
- By applying the general perturbation method of the sum \(S_n\) in the first step and the distributive law in the second step, we get
\[S_n+ ax^{n+1}=ax^0 + \sum_{0\le k\le n} ax^{k+1}= a + x\sum_{0\le k\le n} ax^k=a + xS_n.\]
- By solving for \(S_n\) we obtain as required
\[S_n=\frac{a-ax^{n+1}}{1-x},\quad\quad (x\neq 1).\]
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References
Bibliography
- Graham L. Ronald, Knuth E. Donald, Patashnik Oren: "Concrete Mathematics", Addison-Wesley, 1994, 2nd Edition
