Sum of Sequence of Fourth Powers

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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 using Binomial Coefficients

By definition of binomial coefficient:

\(\ds \binom k 4\) \(=\) \(\ds \frac {k \paren {k - 1} \paren {k - 2} \paren {k - 3} } {4 !}\)
\(\ds \binom k 3\) \(=\) \(\ds \frac {k \paren {k - 1} \paren {k - 2} } {3 !}\)
\(\ds \binom k 2\) \(=\) \(\ds \frac {k \paren {k - 1} } {2 !}\)
\(\ds \binom k 1\) \(=\) \(\ds k\)
\(\ds \leadsto \ \ \) \(\ds 24 \binom k 4\) \(=\) \(\ds k^4 - 6 k^3 + 11 k^2 - 6 k\)
\(\ds 36 \binom k 3\) \(=\) \(\ds 6 k^3 - 18 k^2 + 12 k\)
\(\ds 14 \binom k 2\) \(=\) \(\ds 7 k^2 - 7 k\)
\(\ds \leadsto \ \ \) \(\ds k^4\) \(=\) \(\ds 24 \binom k 4 + 36 \binom k 3 + 14 \binom k 2 + \binom k 1\)


So:

\(\ds \sum_{k \mathop = 0}^n k^4\) \(=\) \(\ds \sum_{k \mathop = 0}^n \paren {24 \binom k 4 + 36 \binom k 3 + 14 \binom k 2 + \binom k 1}\)
\(\ds \) \(=\) \(\ds 24 \binom {n + 1} 5 + 36 \binom {n + 1} 4 + 14 \binom {n + 1} 3 + \binom {n + 1} 2\) Sum of Binomial Coefficients over Upper Index
\(\ds \) \(=\) \(\ds 24 \frac {\paren {n + 1} n \paren {n - 1} \paren {n - 2} \paren {n - 3} } {5 !}\) Definition of Binomial Coefficient
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 36 \frac {\paren {n + 1} n \paren {n - 1} \paren {n - 2} } {4 !}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 14 \frac {\paren {n + 1} n \paren {n - 1} } {3 !}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac {\paren {n + 1} n } {2 !}\)
\(\ds \) \(=\) \(\ds \paren {n + 1} n \paren {\frac {\paren {n - 1} \paren {n - 2} \paren {n - 3} } 5 + \frac {3 \paren {n - 1} \paren {n - 2} } 2 + \frac {7 \paren {n - 1} } 3 + \frac 1 2}\)
\(\ds \) \(=\) \(\ds \frac {\paren {n + 1} n} {30} \paren {6 \paren {n^3 - 6 n^2 + 11 n - 6 } + 45 \paren {n^2 - 3 n + 2} + 70 \paren {n - 1} + 15}\)
\(\ds \) \(=\) \(\ds \frac {\paren {n + 1} n} {30} \paren {6 n^3 - 36 n^2 + 66 n - 36 + 45 n^2 - 135 n + 90 + 70 n - 70 + 15}\)
\(\ds \) \(=\) \(\ds \frac {\paren {n + 1} n} {30} \paren {6 n^3 + 9 n^2 + n - 1}\)
\(\ds \) \(=\) \(\ds \frac {\paren {n + 1} n} {30} \paren {2 n + 1} \paren {3 n^2 + 3 n - 1}\)
\(\ds \) \(=\) \(\ds \frac {\paren {n + 1} n \paren {n + \frac 1 2} \paren {3 n^2 + 3 n - 1} } {15}\)

$\blacksquare$


Proof using Bernoulli Numbers

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 presented as

This result can also be presented as:

$\ds \sum_{k \mathop = 0}^n k^4 = \dfrac {n \paren {n + 1} \paren {2 n + 1} \paren {3 n^2 + 3 n - 1} } {30}$


Also see

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