Sum of Geometric Sequence/Examples/Common Ratio 1
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Theorem
Consider the Sum of Geometric Sequence defined on the standard number fields for all $x \ne 1$.
- $\ds \sum_{j \mathop = 0}^n a x^j = a \paren {\frac {1 - x^{n + 1} } {1 - x} }$
When $x = 1$, the formula reduces to:
- $\ds \sum_{j \mathop = 0}^n a 1^j = a \paren {n + 1}$
Proof
When $x = 1$, the right hand side is undefined:
- $a \paren {\dfrac {1 - 1^{n + 1} } {1 - 1} } = a \dfrac 0 0$
However, the left hand side degenerates to:
\(\ds \sum_{j \mathop = 0}^n a 1^j\) | \(=\) | \(\ds \sum_{j \mathop = 0}^n a\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a \paren {n + 1}\) |
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: Exercise $11$