Can we prove summation formula for the first $n$ terms of natural numbers through calculus? Can we prove summation formula for the first $n$ terms of natural numbers through calculus?
What about the summation of first $n$ numbers of the form $a^k$ and other summation formulas like sum of a GP or AGP?

Essentially using calculus please.
 A: For the sort of arithmetic progression:
If you consider some function $f_k(x)$ that satisfies


*

*$f_k(x)=f(x-1)+x^k$

*$f_k(1)=1$
Then for all $x\in\mathbb N$, $f(x)=\sum_{n=1}^xn^k$.
$$f_k(x)=f_k(x-1)+x^k$$
$$f'_k(x)=f'_k(x-1)+kx^{k-1}$$
$$f'_k(x)=f'_k(x-2)+k\left[(x-1)^{k-1}+x^{k-1}\right]\\\vdots\\ f'_k(x)=f'_k(0)+k\sum_{n=1}^xn^{k-1}\tag{$x\in\mathbb N$}$$
$$f'_k(x)=f'_k(0)+kf_{k-1}(x)$$
$$\begin{align}f_k(x)-\require{cancel}\cancelto0{f_k(0)}&=\int_0^xf'_k(t)dt\\&=\int_0^xf'_k(0)+kf_{k-1}(t)dt\\&=f'(0)x+k\int_0^xf_{k-1}(t)dt\end{align}$$
which sets a recursive formula for finding $f_k(x)$.  Since it is relatively clear that $f_0(x)=x$, one can derive $f_k(x)$ for all $k\in\mathbb N$.  To find $f'_k(0)$, plug in $x=1$.

$$\sum_{n=0}^xr^n=\frac{1-r^{x+1}}{1-r}$$
$$f_0(x)=\lim_{r\to1}\frac{1-r^{x+1}}{1-r}$$
$$\frac{d}{dr}\sum_{n=0}^xr^n=\sum_{n=0}^xnr^{n-1}=\frac{d}{dr}\frac{1-r^{x+1}}{1-r}\tag{AGP?}$$
$$f_1(x)=\lim_{r\to1}\frac{d}{dr}\frac{1-r^{x+1}}{1-r}$$
In general,
$$f_k(x)=\lim_{r\to1}\underbrace{\left(\frac{d}{dr}r\left(\frac{d}{dr}r\left(\dots r\left(\frac{d}{dr}\frac{1-r^{x+1}}{1-r}\right)\dots\right)\right)\right)}_{k\ \frac{d}{dr}'s}$$

For the geometric progression:
$$\frac{1-r^{x+1}}{1-r}=\frac1{1-r}-\frac{r^{x+1}}{1-r}$$
Take the taylor expansion of both expressions around $r=0$ and you will get
$$=\left(1+r+r^2+\dots+r^x+\color{red}{r^{x+1}+\dots}\right)-\left(\color{red}{r^{x+1}+r^{x+2}+\dots}\right)$$
$$=1+r+r^2+\dots+r^x$$
Doing this backwards would result in solving a differential equation at the step where we take the taylor expansion.
A: Well if you mean $a^1+a^2+...+a^n$, just plug in the arithmetic progression formula. If you want to calculate $1^k+2^k+...+n^k$, here is an iterative method that you may apply.
For instance, when we are calculating $n^3$, in fact $(n+3)(n+2)(n+1)n-(n+2)(n+1)n(n-1)=4(n+2)(n+1)n=4(n^3+3n^2+2n)$
$$\sum_{i=1}^n (n+3)(n+2)(n+1)n-(n+2)(n+1)n(n-1)=\sum_{i=1}^n 4(n^3+3n^2+2n)$$
We can subtract the sum of $n^2$ and $n$ part to get the sum of $n^3$ part.
