Proposition: Sum of Squares
The sum of consecutive square numbers to a given number $n^2$ can be calculated by the following formula
$$\sum_{k=0}^n k^2=\frac{n(n+1)(2n+1)}6.$$
Table of Contents
Proofs: 1
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References
Bibliography
- Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983
