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

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