# Sum of Sequence of Fourth Powers

## 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

 $\ds \sum_{i \mathop = 1}^n i^p$ $=$ $\ds 1^p + 2^p + \cdots + n^p$ $\ds$ $=$ $\ds \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}$

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:

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