<s>[INST] What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression using sums $\sum$ and double sums $\sum$$\sum$?
 [/INST] The sum can be derived via combinatorial argument.
Count the number of quintuples $(x_1,x_2,x_3, x_4,x_5)$ with the property $x_5> \max\limits_{i\in S}$ from the set $S=\{1,2,...,n,n+1\}$,
then as $x_5$ varies from $2$ to $n+1$, the positions $(x_1,x_2,x_3, x_4)$
 the positions can be filled in $\sum\limits_{i=1}^n i^4$ ways.
Counting it according to the cases: $5 = 4 + 1 = 2 + 2 + 1 = 3 + 1 + 1 = 2+1+1+1=1+1+1+1+1$ (where the numbers in partitions indicate the number of equal components in that partition, i.e. for example $5=4+1$ corresponds to the case $x_1=x_2=x_3=x_4<x_5$; and $2+2+1$ corresponds to the case two pairs have equal value but the value of the pair are different ,.. and so on). Count the number of ways in which each of the partitioning can happen: $1,14,36,24$ respectively (in the same order as I have presented the partitions).
You get the number of ways to be: ${n+1\choose 2}+14{n+1\choose 3}+36{n+1 \choose 4}+24{n+1\choose 5}$
Therefore, $\sum\limits_{i=1}^n i^4 = {n+1\choose 2}+14{n+1\choose 3}+36{n+1 \choose 4}+24{n+1\choose 5} = (1/5)n^5 + (1/2)n^4 + (1/3)n^3 - (1/30)n$.
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