Sum of Sequence of Fourth Powers/Proof using Bernoulli Numbers

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \sum_{k \mathop = 0}^n k^4 = \dfrac {\paren {n + 1} n \paren {n + \frac 1 2} \paren {3 n^2 + 3 n - 1} } {15}$


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 $p = 4$:

\(\ds \sum_{i \mathop = 1}^n i^4\) \(=\) \(\ds \frac {n^{4 + 1} } {4 + 1} + \sum_{k \mathop = 1}^4 \frac {B_k \, 4^{\underline {k - 1} } \, n^{4 - k + 1} } {k!}\)
\(\ds \) \(=\) \(\ds \frac {n^5} 5 + \frac {B_1 \, 4^{\underline 0} \, n^4} {1!} + \frac {B_2 \, 4^{\underline 1} \, n^3} {2!} + \frac {B_3 \, 4^{\underline 2} \, n^2} {3!} + \frac {B_4 \, 4^{\underline 3} \, n} {4!}\)
\(\ds \) \(=\) \(\ds \frac {n^5} 5 + \frac 1 2 \frac {n^4} {1!} + \frac 1 6 \frac {4 n^3} {2!} + 0 \frac {4 \times 3 n^2} {3!} - \frac 1 {30} \frac {4 \times 3 \times 2 n} {4!}\) Definition of Bernoulli Numbers and Definition of Falling Factorial
\(\ds \) \(=\) \(\ds \frac {n^5} 5 + \frac {n^4} 2 + \frac {n^3} 3 - \frac n {30}\) simplifying
\(\ds \) \(=\) \(\ds \frac {6 n^5 + 15 n^4 + 10 n^3 - n} {30}\)
\(\ds \) \(=\) \(\ds \frac {n \paren {n + 1} \paren {2 n + 1} \paren {3 n^2 + 3 n - 1} } {30}\)


Hence the result.

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Sum_of_Sequence_of_Fourth_Powers/Proof_using_Bernoulli_Numbers&oldid=514065"