I asked this question here
When I was in high school, I was fascinated by
$$
\sum\limits_{k=1}^n k= \frac{n(n+1)}{2}
$$ so I tried to find the general value of the sum
$$
\sum\limits_{k=1}^n k^i\;\text{ for all }\; i \in \mathbb{N}
$$$$
\sum\limits_{k=1}^n k^m\;\text{ for all }\; m \in \mathbb{N}
$$
I was able to to find the sum up to $i=6$$m=6$. Here, I tried to search for a pattern to find the general solution of $\sum\limits_{k=1}^n k^i $$\sum\limits_{k=1}^n k^m $ s.t $i \in \mathbb{N}$$m \in \mathbb{N}$ which I failed to do, but I noticed this pattern:
$$
\begin{align}
S_1(n) & :={\sum\limits_{k=1}^n k= \frac{n(n+1)}{2}}\\
S_2(n) & :={\sum\limits_{k=1}^n k^2= \color{blue}{\frac{n(2n+1)(n+1)}{6}}}\\
S_3(n) & :={\sum\limits_{k=1}^n k^3= \color{red}{\frac{\color{red}{n^2(n+1)^2}}{\color{red}{4}}}}\\
S_4(n) & :={\sum\limits_{k=1}^n k^4=\color{blue}{\frac{n(2n+1)(n+1)}{6}} \cdot \frac{3n^2+3n-1}{ 5}}\\
S_5(n) & :={\sum\limits_{k=1}^n k^5= \color{red}{\frac{\color{red}{n^2(n+1)^2}}{\color{red}{4}}} \cdot\frac{2n^2+2n-1}{ 3}}\\
S_6(n) & :={\sum\limits_{k=1}^n k^6= \color{blue}{\frac{n(2n+1)(n+1)}{6}} \cdot \frac{3n^4+6n^3-3n+1}{ 7 }}\\
S_7(n) & :={\sum\limits_{k=1}^n k^7= \color{red}{\frac{\color{red}{n^2(n+1)^2}}{\color{red}{4}} }\cdot\frac{3n^4+6n^3-n^2-4n+2}{ 6}}\\
S_8(n) &:= {\sum\limits_{k=1}^n k^8= \color{blue}{\frac{n(2n+1)(n+1)}{6}} \cdot \frac{5n^6+15n^5+5n^4-15n^3-n^2+9n+3}{ 15}}\\
S_9(n) &:= {\sum\limits_{k=1}^n k^9= \color{red}{\frac{\color{red}{n^2(n+1)^2}}{\color{red}{4}}} \cdot\frac{(n^2+n-1)(2n^4 +4n^3-n^3-3n^2+3)}{ 5}}\\
S_{10}(n) & := {\sum\limits_{k=1}^n k^{10}= \color{blue}{\frac{n(2n+1)(n+1)}{6}} \cdot \frac{ 3 n^8+ 12 n^7+ 8 n^6 - 18 n^5- 10 n^4+ 24 n^3 + 2 n^2 - 15 n +5}{ 11}}\\
S_{11}(n) & :={\sum\limits_{k=1}^n k^{11}= \color{red}{\frac{\color{red}{n^2(n+1)^2}}{\color{red}{4}}} \cdot\frac{2n^8 +8n^7+4n^6-16n^5-5n^4+26n^3-3n^2-20n+10}{ 6}}
\end{align}
$$
I noticed that:
- For odd $i>1$$m>1$, $$ \sum\limits_{k=1}^n k^i= \color{red}{\frac{{n^2(n+1)^2}}{{4}}} \cdot \frac{P_{i-3}(n)}{N_i} $$$$ \sum\limits_{k=1}^n k^m= \color{red}{\frac{{n^2(n+1)^2}}{{4}}} \cdot \frac{P_{m-3}(n)}{N_m} $$ s.t $P_{i-3}(n)$$P_{m-3}(n)$ is an $i-3$$m-3$ polynomial with integer coefficients $ \{a_{i-3},\ldots a_1,a_0 \}$$ \{a_{m-3},\ldots a_1,a_0 \}$ such that $\gcd \{a_{i-3},\ldots a_1,a_0 \}=1$$\gcd \{a_{m-3},\ldots a_1,a_0 \}=1$, $N_i\in \mathbb {N}$$N_m\in \mathbb {N}$.
- For even $i$$m$, $$ \sum\limits_{k=1}^n k^i= \color{blue}{\frac{n(2n+1)(n+1)}{6}}\cdot \frac{P_{i-2}'(n)}{N_i} $$$$ \sum\limits_{k=1}^n k^m= \color{blue}{\frac{n(2n+1)(n+1)}{6}}\cdot \frac{P_{m-2}'(n)}{N_m} $$ s.t $P_{i-2}'(n)$$P_{m-2}'(n)$ is an $i-2$$m-2$ polynomial with integer coefficients $\{ a_{i-2},\dots a_1,a_0 \}$$\{ a_{m-2},\dots a_1,a_0 \}$ such that $\gcd \{ a_{i-2},\dots a_1,a_0 \}=1$$\gcd \{ a_{m-2},\dots a_1,a_0 \}=1$, $N_i\in \mathbb {N}$$N_m\in \mathbb {N}$.
When I was in high school I couldn't prove this pattern, and I remembered this observation that I had totally forgotten about. Now, after two years from my first attempt, I tried to prove this pattern again, but I couldn't.
This question has been partly answered here (The answer shows that $\displaystyle\sum\limits_{k=1}^n k^i$$\displaystyle\sum\limits_{k=1}^n k^m$ is divisible by ${n^2(n+1)^2}$ for odd $i>1$$m>1$ and $\displaystyle\sum\limits_{k=1}^n k^i$$\displaystyle\sum\limits_{k=1}^n k^m$ is divisible by ${n(2n+1)(n+1)}$ for even $i$$m$) the only missing part is to show that the denominator is a multiple of $4$ if $i\in 2\mathbb{N}+1$$m\in 2\mathbb{N}+1$, and the denominator is a multiple of $6$ if $i \in 2\mathbb{N}$$m \in 2\mathbb{N}$.
Update With the help of the paper that @SamHopkins recommended I posted an answer here
