Sum of Sequence of Squares/Proof using Bernoulli Numbers
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Theorem
- $\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$
Proof
From Sum of Powers of Positive Integers:
\(\ds \sum_{i \mathop = 1}^n i^p\) | \(=\) | \(\ds 1^p + 2^p + \cdots + n^p\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}\) |
where $B_k$ are the Bernoulli numbers.
Setting $p = 2$:
\(\ds \sum_{i \mathop = 1}^n i^2\) | \(=\) | \(\ds 1^2 + 2^2 + \cdots + n^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^{2 + 1} } {2 + 1} + \frac {B_1 \, p^{\underline 0} \, n^2} {1!} + \frac {B_2 \, p^{\underline 1} \, n^1} {2!}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^3} 3 + B_1 n^2 + B_2 n\) | Definition of Falling Factorial and simplification | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^3} 3 + \frac {n^2} 2 + \frac n 6\) | Definition of Bernoulli Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n \paren {n + 1} \paren {2 n + 1} } 6\) | after algebra |
Hence the result.
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 19$: Sums of Powers of Positive Integers: $19.10$