Sum of Sequence of Fourth Powers/Proof using Binomial Coefficients

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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

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
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 36 \frac {\paren {n + 1} n \paren {n - 1} \paren {n - 2} } {4 !}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 14 \frac {\paren {n + 1} n \paren {n - 1} } {3 !}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac {\paren {n + 1} n } {2 !}\)
\(\ds \) \(=\) \(\ds \paren {n + 1} n \paren {\frac {\paren {n - 1} \paren {n - 2} \paren {n - 3} } 5 + \frac {3 \paren {n - 1} \paren {n - 2} } 2 + \frac {7 \paren {n - 1} } 3 + \frac 1 2}\)
\(\ds \) \(=\) \(\ds \frac {\paren {n + 1} n} {30} \paren {6 \paren {n^3 - 6 n^2 + 11 n - 6 } + 45 \paren {n^2 - 3 n + 2} + 70 \paren {n - 1} + 15}\)
\(\ds \) \(=\) \(\ds \frac {\paren {n + 1} n} {30} \paren {6 n^3 - 36 n^2 + 66 n - 36 + 45 n^2 - 135 n + 90 + 70 n - 70 + 15}\)
\(\ds \) \(=\) \(\ds \frac {\paren {n + 1} n} {30} \paren {6 n^3 + 9 n^2 + n - 1}\)
\(\ds \) \(=\) \(\ds \frac {\paren {n + 1} n} {30} \paren {2 n + 1} \paren {3 n^2 + 3 n - 1}\)
\(\ds \) \(=\) \(\ds \frac {\paren {n + 1} n \paren {n + \frac 1 2} \paren {3 n^2 + 3 n - 1} } {15}\)

$\blacksquare$


Also see

Sources