# Can we prove summation formula for the first $n$ terms of natural numbers through calculus? [closed]

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?

## closed as unclear what you're asking by Jack, Rob Arthan, Claude Leibovici, user91500, WatsonOct 16 '16 at 8:46

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• You need difference equations, analogous to—but different from—derivatives. – Ted Shifrin Oct 16 '16 at 0:53
• See Faulhaber's formula. You can search this site for it as well. – Ross Millikan Oct 16 '16 at 0:55
• Do you mean $1^k+2^k+\dots+n^k$ or do you mean $a^1+a^2+\dots+a^n$? – Simply Beautiful Art Oct 16 '16 at 1:09
• @SimpleArt both – ankit Oct 16 '16 at 1:11
• For $1^k+2^k+\dots+n^k$. – Simply Beautiful Art Oct 16 '16 at 1:27

For the sort of arithmetic progression:

If you consider some function $f_k(x)$ that satisfies

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

2. $f_k(1)=1$

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.

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.

• Um, you mean geometric progression in the first line? And the OP seems to know this and wants this done with calculus. – Simply Beautiful Art Oct 16 '16 at 1:28
• Yes as Simple Art says I know it. that is why I tagged it with "alternative-proof" – ankit Oct 16 '16 at 2:13