# 数学代写|组合学代写Combinatorics代考|The Bernoulli and Cauchy numbers

#### Doug I. Jones

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## 数学代写|组合学代写Combinatorics代考|Power sums

The Stirling numbers can be used to calculate sums of powers of consecutive integer numbers. Let us first recall the well-known formulas from elementary mathematics:
$$1+2+3+\cdots+n=\frac{1}{2}\left(n^2+n\right)$$
The sum of the squares of the first $n$ numbers is
$$1^2+2^2+3^2+\cdots+n^2=\frac{1}{6}\left(2 n^3+3 n^2+n\right) .$$
For the third powers we have that
$$1^3+2^3+3^3+\cdots+n^3=\frac{1}{4}\left(n^4+2 n^3+n^2\right) .$$
We can easily recognize a rule: the sums of the powers of the first $n$ positive integers can be expressed as polynomials of $n$. It would be interesting to know what are the coefficients of these polynomials in general. The sums on the left-hand sides will be called power sums.
The Stirling numbers pop up in this problem via the relation (2.46):
$$x^p=\sum_{k=0}^p\left{\begin{array}{l} p \ k \end{array}\right} x^{k}$$
On the left-hand side there is the $p$ th power of a real number $x$. Sum on $x$ from 1 to $n-1$. (We shall see later that summing up to $n-1$ in place of $n$ makes the resulting expressions somewhat simpler looking.)
$$1^p+2^p+\cdots+(n-1)^p=\sum_{x=1}^{n-1} \sum_{k=0}^p\left{\begin{array}{l} p \ k \end{array}\right} x^{k} .$$
If we could find a simple polynomial expression for the right-hand side we would be ready. What we immediately see is that we should know the sums of the falling factorials, since
$$\sum_{x=1}^{n-1} \sum_{k=0}^p\left{\begin{array}{l} p \ k \end{array}\right} x^{k}=\sum_{k=0}^p\left{\begin{array}{l} p \ k \end{array}\right} \sum_{x=1}^{n-1} x^{k}$$

## 数学代写|组合学代写Combinatorics代考|Power sums of arithmetic progressions

The power sum formula (5.5) can easily be generalized so that we are summing arithmetic progressions’ powers. Arithmetic sequences are of the form $r, r+m, r+2 m, \ldots, r+n m$. We will prove very briefly that
$$\sum_{i=1}^n(r+i m)^p=m^p \sum_{k=1}^{p+1}(k-1) !\left{\begin{array}{c} p+1 \ k \end{array}\right}\left[\left(\begin{array}{c} n+r / m \ k \end{array}\right)-\left(\begin{array}{c} r / m \ k \end{array}\right)\right],$$
given that $m$ is not zero ${ }^2$.
Applying (5.5) with $n=n+t$ and $n=t$, and subtracting the second expression from the first, it comes that
$$\sum_{i=1}^n(t+i)^p=\sum_{k=1}^{p+1}(k-1) !\left{\begin{array}{c} p+1 \ k \end{array}\right}\left[\left(\begin{array}{c} n+t \ k \end{array}\right)-\left(\begin{array}{l} t \ k \end{array}\right)\right] .$$
As this formula is valid for all positive integer $t$, and on both sides we have polynomials of $t$, it follows that the formula is valid for all real $t$. Let us write $t$ as a rational number $t=r / m$. Then multiply both sides of the last formula with $m^p$. We get that
$$\sum_{i=1}^n(r+i m)^p=m^p \sum_{k=1}^{p+1}(k-1) !\left{\begin{array}{c} p+1 \ k \end{array}\right}\left[\left(\begin{array}{c} n+r / m \ k \end{array}\right)-\left(\begin{array}{c} r / m \ k \end{array}\right)\right] .$$
This is exactly the formula that we wanted to verify.

# 组合学代考

## 数学代写|组合学代写Combinatorics代考|Power sums

$$1+2+3+\cdots+n=\frac{1}{2}\left(n^2+n\right)$$

$$1^2+2^2+3^2+\cdots+n^2=\frac{1}{6}\left(2 n^3+3 n^2+n\right) .$$

$$1^3+2^3+3^3+\cdots+n^3=\frac{1}{4}\left(n^4+2 n^3+n^2\right) .$$

## 数学代写|组合学代写Combinatorics代考|Power sums of arithmetic progressions

$r, r+m, r+2 m, \ldots, r+n m$. 我们将非常简要地 证明

Isum_{i=1 {}$^{\wedge} n(t+i)^{\wedge} p=$ Isum_ ${k=1}^{\wedge}{p+1}(k-1) ! \backslash$ left ${\backslash$ begin {

Isum_{i=1 $}^{\wedge} n(r+i m)^{\wedge} p=m^{\wedge} p$ Isum_{k=1 $}^{\wedge}{p+1}(k-1) ! \backslash$ left

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