How to write the sum of r^(α+1) in terms of the sum of r^(α-1)?

Question

The sum from 1 to n of r^(α+1), {α,n} $\in ℕ $, can be written as the sum from 1 to n of the sum from x to n of r^α.

$$\underset{r = 1}{\overset{n}{\sum }}r^{\alpha + 1} = \underset{x = 1}{\overset{n}{\sum }}\underset{r = x}{\overset{n}{\sum }}r^\alpha$$ How can one write the sum from 1 to n of r^(α+1) in terms of anything less than α, since trying to write the sum of a sum of a sum would be using the same variable x repeatedly.

Main question: How can one write $\underset{r = 1}{\overset{n}{\sum }}r^{\alpha + 1}$ in terms of $r^{\alpha - 1}$.

Then: How can one write it more generally, in terms of $r^{\beta}$, where $\alpha > \beta$ and $\beta \in ℕ $

In case the above didn't make sense, an example: $$\underset{r = 1}{\overset{n}{\sum }}r^4 = \underset{x = 1}{\overset{n}{\sum }}\underset{r = x}{\overset{n}{\sum }}r^3$$

How does one write $\underset{r = 1}{\overset{n}{\sum }}r^4$ in terms of $\underset{r = 1}{\overset{n}{\sum }}r^2$ or $\underset{r = 1}{\overset{n}{\sum }}r$, as substituting ($\underset{r = 1}{\overset{n}{\sum }}r^3 = \underset{x = 1}{\overset{n}{\sum }}\underset{r = x}{\overset{n}{\sum }}r^2$) into $\underset{r = 1}{\overset{n}{\sum }}r^4$ doesn't work directly, since the variable x is used more than once. $$\underset{r = 1}{\overset{n}{\sum }}r^4 \neq \underset{x = 1}{\overset{n}{\sum }}\underset{x = 1}{\overset{n}{\sum }}\underset{r = x}{\overset{n}{\sum }}r^2$$ Changing the second summation's bounds also doesn't work to fix the problem.

I realised that putting brackets around the substituted sums will solve the problem, but could it be simplified further in terms of summations of summations of summations... instead of putting brackets around the substituted in equivalences of lower and lower powers. Can the summation be altered to compensated for the substituted summation of summations instead?

^That's some what true, but the problem of variables remain even if you used brackets, only just remembered why I ruled out that answer in this 2nd edit

Using sum of differences can give you the general formula of summations of natural numbers to natural powers, but how could we express the general formulae of an arbitrarily large power in terms of any smaller powers.

Answer 1

Here's a way of generating the successive formulas for the sums of powers in terms of the sums of previous powers I came up with a while ago. This is obviously not original, and was probably known to Bernoulli, but it was found independently by me.

If $f_k(n) = \sum_{i=1}^n i^k$ ($f_0(n) = n$) then

$f_{k}(n) =\dfrac1{k+1}\left((n+1)^{k+1}-1-\sum_{j=0}^{k-1} \binom{k+1}{j}f_j(n)\right) $

Proof.

$\begin{array}\\ f_k(n+1) &= \sum_{i=1}^{n+1} i^k\\ &= \sum_{i=0}^{n} (i+1)^k\\ &=1+ \sum_{i=1}^{n} (i+1)^k\\ &=1+ \sum_{i=1}^{n} \sum_{j=0}^k \binom{k}{j}i^j\\ &=1+ \sum_{j=0}^k \binom{k}{j}\sum_{i=1}^{n}i^j\\ &=1+ \sum_{j=0}^k \binom{k}{j}f_j(n)\\ \text{so}\\ (n+1)^k &=f_{k}(n+1)-f_k(n)\\ &=1+ \sum_{j=0}^k \binom{k}{j}f_j(n)-f_k(n)\\ &=1+ \sum_{j=0}^{k-1} \binom{k}{j}f_j(n)\\ \text{or}\\ kf_{k-1}(n) &=(n+1)^k-1-\sum_{j=0}^{k-2} \binom{k}{j}f_j(n)\\ \text{or}\\ (k+1)f_{k}(n) &=(n+1)^{k+1}-1-\sum_{j=0}^{k-1} \binom{k+1}{j}f_j(n)\\ \text{or}\\ f_{k}(n) &=\dfrac1{k+1}\left((n+1)^{k+1}-1-\sum_{j=0}^{k-1} \binom{k+1}{j}f_j(n)\right)\\ \end{array} $

Note that this shows that if $f_j(n)$ is a polynomial of degree $j+1$ for $0 \le j \le k-1$ then $f_k(n)$ is a polynomial of degree $k+1$.

Here are the first few cases.

$\begin{array}\\ \\ k=1\\ 2f_{1}(n) &=(n+1)^{2}-1-f_0(n)\\ &=n^2+2n+1-1-n\\ &=n^2+n\\ f_1(n) &=\dfrac{n^2+n}{2}\\ \\ k=2\\ 3f_{2}(n) &=(n+1)^{3}-1-f_0(n)-3f_1(n)\\ &=n^3+3n^2+3n+1-1-n-3(n^2+n)/2\\ &=\dfrac{2n^3+6n^2+6n+2-2-2n-3(n^2+n)}{2}\\ &=\dfrac{2n^3+3n^2+n}{2}\\ &=\dfrac{n(2n+1)(n+1)}{2}\\ f_{2}(n) &=\dfrac{n(2n+1)(n+1)}{6}\\ &=\dfrac{2n^3+3n^2+n}{6}\\ \\ k=3\\ 4f_{3}(n) &=(n+1)^{4}-1-f_0(n)-4f_1(n)-6f_2(n)\\ &=n^4+4n^3+6n^2+4n+1-1-n-4\dfrac{n^2+n}{2}-6\dfrac{n(2n+1)(n+1)}{6}\\ &=n^4+4n^3+6n^2+3n-2(n^2+n)-(2n^3+3n^2+n)\\ &=n^4+2n^3+n^2\\ &=n^2(n^2+2n+1)\\ &=n^2(n+1)^2\\ f_3(n) &=\dfrac{n^2(n+1)^2}{4}\\ &=\left(\dfrac{n(n+1)}{2}\right)^2\\ \\ k=4\\ 5f_{4}(n) &=(n+1)^{5}-1-f_0(n)-5f_1(n)-10f_2(n)-10f_3(n)\\ &=n^5+5n^4+10n^3+10n^2+5n+1-1-n-5\dfrac{n^2+n}{2}-10\dfrac{n(2n+1)(n+1)}{6}-10\left(\dfrac{n(n+1)}{2}\right)^2\\ &=n^5 + \dfrac{5 n^4}{2} + \dfrac{5 n^3}{3} - \dfrac{n}{6}\\ &=\dfrac{6n^5 + 15 n^4 + 10 n^3 - n}{6}\\ &=\dfrac{ n (2 n + 1) (n + 1) (3 n^2 + 3 n - 1)}{6}\\ f_4(n) &=\dfrac{ n (2 n + 1) (n + 1) (3 n^2 + 3 n - 1)}{30}\\ &=\dfrac{6n^5 + 15 n^4 + 10 n^3 - n}{30}\\ &=\dfrac{n^5}{5} + \dfrac{ n^4}{2} + \dfrac{ n^3}{3} - \dfrac{n}{30}\\ \end{array} $

Continuing this, here is a first step to discovering the Bernoulli numbers.

If $f_j(n) =\sum_{i=1}^{j+1} c_{j, i} n^i $, then

$\begin{array}\\ c_{0, 1..1} &=1\\ c_{1, 1..2} &=\dfrac12. \dfrac12\\ c_{2, 1..3} &=\dfrac16. \dfrac12, \dfrac13\\ c_{3, 1..4} &=0, \dfrac14. \dfrac12, \dfrac14\\ c_{4, 1..5} &=- \dfrac1{30}, 0, \dfrac13. \dfrac12, \dfrac15\\ \end{array} $

and

$\begin{array}\\ (k+1)f_{k}(n) &=(n+1)^{k+1}-1-\sum_{j=0}^{k-1} \binom{k+1}{j}f_j(n)\\ &=\sum_{j=0}^{k+1}\binom{k+1}{j}n^j-1-\sum_{j=0}^{k-1} \binom{k+1}{j}\sum_{i=1}^{j+1} c_{j, i} n^i\\ &=\sum_{j=1}^{k+1}\binom{k+1}{j}n^j-\sum_{i=1}^{k}\sum_{j=i-1}^{k-1} \binom{k+1}{j} c_{j, i} n^i\\ &=\sum_{i=1}^{k+1}\binom{k+1}{i}n^i-\sum_{i=1}^{k}n^i\sum_{j=i-1}^{k-1} \binom{k+1}{j} c_{j, i}\\ &=n^{k+1}+\sum_{i=1}^{k}\binom{k+1}{i}n^i-\sum_{i=1}^{k}n^i\sum_{j=i-1}^{k-1} \binom{k+1}{j} c_{j, i}\\ &=n^{k+1}+\sum_{i=1}^{k}n^i\left(\binom{k+1}{i}-\sum_{j=i-1}^{k-1} \binom{k+1}{j} c_{j, i}\right)\\ \text{so}\\ (k+1)c_{k, k+1} &=1\\ c_{k, k+1} &=\dfrac1{k+1}\\ \text{for } 1 \le i \le k\\ (k+1)c_{k, i} &=\binom{k+1}{i}-\sum_{j=i-1}^{k-1} \binom{k+1}{j} c_{j, i}\\(k+1)c_{k, k} &=\binom{k+1}{k}-\sum_{j=k-1}^{k-1} \binom{k+1}{j} c_{j, k}\\ &=k+1-\binom{k+1}{k-1} c_{k-1, k}\\ &=k+1-\dfrac{k(k+1)}{2}\dfrac1{k}\\ &=k+1-\dfrac{k+1}{2}\\ &=\dfrac{k+1}{2}\\ c_{k, k} &=\dfrac12\\ \\ (k+1)c_{k, i} &=\binom{k+1}{i}-\sum_{j=i-1}^{k-1} \binom{k+1}{j} c_{j, i}\\(k+1)c_{k, k-1} &=\binom{k+1}{k-1}-\sum_{j=k-2}^{k-1} \binom{k+1}{j} c_{j, k-1}\\ &=\dfrac{(k+1)k}{2}-\binom{k+1}{k-2} c_{k-2, k-1}-\binom{k+1}{k-1} c_{k-1, k-1}\\ &=\dfrac{(k+1)k}{2}-\dfrac{(k+1)k(k-1)}{6} c_{k-2, k-1}-\dfrac{(k+1)k}{2}\dfrac12\\ &=\dfrac{(k+1)k}{4}-\dfrac{(k+1)k(k-1)}{6} c_{k-2, k-1}\\ c_{k, k-1} &=\dfrac{k}{4}-\dfrac{k(k-1)}{6} c_{k-2, k-1}\\ &=\dfrac{k}{4}-\dfrac{k(k-1)}{6} \dfrac1{k-1}\\ &=\dfrac{k}{4}-\dfrac{k}{6}\\ &=\dfrac{k}{12}\\ \\ (k+1)c_{k, i} &=\binom{k+1}{i}-\sum_{j=i-1}^{k-1} \binom{k+1}{j} c_{j, i}\\(k+1)c_{k, k-2} &=\binom{k+1}{k-2}-\sum_{j=k-3}^{k-1} \binom{k+1}{j} c_{j, k-2}\\ &=\binom{k+1}{k-2}- \binom{k+1}{k-3} c_{k-3, k-2}-\binom{k+1}{k-2} c_{k-2, k-2}-\binom{k+1}{k-1} c_{k-1, k-2}\\ &=\dfrac{(k+1)k(k-1)}{6}- \dfrac{(k+1)k(k-1)(k-2)}{24}\dfrac1{k-2}-\dfrac{(k+1)(k(k-1)}{6}\dfrac12-\dfrac{(k+1)k}{2}\dfrac{k-1}{12}\\ &=\dfrac{(k+1)k(k-1)}{6}- \dfrac{(k+1)k(k-1)}{24}-\dfrac{(k+1)(k(k-1)}{12}-\dfrac{(k+1)k(k-1)}{24}\\ c_{k, k-2} &=\dfrac{k(k-1)}{6}- \dfrac{k(k-1)}{24}-\dfrac{k(k-1)}{12}-\dfrac{k(k-1)}{24}\\ &=0\\ \\ (k+1)c_{k, i} &=\binom{k+1}{i}-\sum_{j=i-1}^{k-1} \binom{k+1}{j} c_{j, i}\\(k+1)c_{k, k-3} &=\binom{k+1}{k-3}-\sum_{j=k-4}^{k-1} \binom{k+1}{j} c_{j, k-3}\\ &=\binom{k+1}{k-3}\\ &- \binom{k+1}{k-4} c_{k-4, k-3}\\ &-\binom{k+1}{k-3} c_{k-3, k-3}\\ &-\binom{k+1}{k-2} c_{k-2, k-3}\\ &-\binom{k+1}{k-1} c_{k-1, k-3}\\ &=\dfrac{(k+1)k(k-1)(k-2)}{24}\\ &- \dfrac{(k+1)k(k-1)(k-2)(k-3)}{120}\dfrac1{k-3}\\ &-\dfrac{(k+1)k(k-1)(k-2)}{24}\dfrac12\\ &-\dfrac{(k+1)k(k-1)}{6}\dfrac{k-2}{12}\\ &-\dfrac{(k+1)k}{2}0\\ &=\dfrac{(k+1)k(k-1)(k-2)}{24}\\ &- \dfrac{(k+1)k(k-1)(k-2)}{120}\\ &-\dfrac{(k+1)k(k-1)(k-2)}{48}\\ &-\dfrac{(k+1)k(k-1)(k-2)}{72}\\ &=(k+1)k(k-1)(k-2)(\dfrac1{24} - \dfrac1{120} -\dfrac1{48} -\dfrac1{72})\\ &=-\dfrac{(k+1)k(k-1)(k-2)}{720}\\ c_{k, k-3} &=-\dfrac{k(k-1)(k-2)}{720}\\ \\ c_{k, k+1} &=\dfrac1{k+1}\\ c_{k, k} &=\dfrac12\\ c_{k, k-j} &=b_j\prod_{i=0}^{j-1}(k-i) \\ &=b_j\dfrac{\prod_{i=0}^{k-1}(k-i)}{\prod_{i=j}^{k-1}(k-i)} \\ &=b_j\dfrac{k!}{\prod_{i=1}^{k-j}(i)} \\ &=b_j\dfrac{k!}{(k-j)!} \\ &=j!b_j\binom{k}{j} \\ b_{1, 2, 3} &=\dfrac1{12}, 0, -\dfrac1{720}\\ b_j(j+1)! &=\dfrac12, 0, -\dfrac1{30}\\ \\ \end{array} $