I wonder if it is possible to compose formulas by hand like: $1+2+3+\cdots+n$. The formula for this is:$(n)(n+1)/2$.
And for $1^2 + 2^2 + 3^2+\cdots+n^2$ you have the formula: ...
But this is more a trial and error work. So my question is, is it possible too generate series of the form: $1^3+2^3+3^3+\cdots+n^3$ or $1^4+2^4+3^4+\cdots+n^4$, etc. by logic and not by trial and error?
Thank you
The formula for $1^2 + 2^2 + 3^2+\cdots+n^2$ is not by trial and error. There is a proof by logic, which is as follows.
Note $$(n+1)^3-1^3=\sum_{k=1}^{n}(k+1)^3-k^3=\sum_{k=1}^{n}(3k^2+3k+1)$$ $$=3\sum_{k=1}^{n} k^2+3\sum_{k=1}^{n} k+\sum_{k=1}^{n} 1$$
This gives us that $$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6} = \frac{2n^3 + 3n^2 + n}{6}$$ We can do something similar for $n^3$, $n^4$, and there is a formuala for the $p$-th powers called Faulhaber's Formula.