SUM OF POWERS OF NATURAL NUMBERS VIA BINOMIAL COEFFICIENTS

In this paper, many properties of several sequences such as Binomial coefficients, the falling numbers, and Stirling numbers of the second kind are discussed. We use these properties to find recurrence equations for the sums of powers σm(n) = ∑k=1 km interm of Binomial coefficients and Stirling numbers.

For further properties of {σ m (n)} ∞ m=0 , see [5] and [8]. In this paper, we are interested in constructing recurrence equations for σ m (n), this will allow us, for example, to find σ m+1 (n) using σ m (n). Also, we will be able to find σ m+2 (n) using σ m+1 (n) and σ m (n). We will give definitions and needed results in subsections 2.1, 2.2, and 2.3.
In section 3, we construct and prove recurrence equations for σ m (n).

PRELIMINARIES
2.1. Binomial coefficients and falling numbers. Let N 0 be the set of nonnegative integers.
For n ∈ N 0 , the falling factorial numbers (x) n , introduced by Leo August Pochhammer, are given as where (x) 0 = 1. It is clear that (x) n = Γ(x+1) Γ(x+1−n) . In particular, for m, n ∈ N 0 The binomial coefficient , denoted by n k is the coefficient of the x k -term in the polynomial expansion of (1 + x) n . It can be found by the formula Clearly, using (2), for the binomial coefficients and falling numbers are related as Therefore, Consequently, Taking the inverse difference operator for both sides of (4), we get (k) n = n∆ −1 (k) n−1 . Consequently, This also gives Following the same arguments of Example 17, one can prove the following result which is needed later.
This proposition leads to the following result.
Also, it is easy to show that the forward difference operator satisfies the following proposition For example, using (7) and for j ∈ N This equations defines the inverse difference operator as For further properties for difference operators and their applications, see [9], [1], [2], [4] and [3].

Stirling numbers.
For n ∈ N 0 , the sequence { n k }, k = 0, 1, 2 · · · n, which satisfies is called the Stirling numbers of the second kind.
In [10], the following explicit formula is given to calculate these numbers This equation implies that . Thus, for m ∈ N and using (14), we have Taking the sum for both sides of (15) from k = 1 to n and using Proposition 2.4 give Example 2.6. Using Example 2.5, we get that For further properties of the Stirling numbers, see [11].

MAIN RESULTS
Proposition 3.1. For m ∈ N 0 , the falling numbers (x) n satisfies Proof. Therefore, Now, we construct recurrence relations for σ m (n). This can be done after proving the following result.
Proposition 3.2. For m, k ∈ N, the following are true Proof. Multiply (12) by k+1 1 and use Proposition 3.1 to get Now, use Proposition 3.1 to get Again, use Proposition 3.1 to get Taking the sum for the results in Proposition 3.2 from k = 1 to n and using (9) to get the following recurrence relations for σ m (n).

CONFLICT OF INTERESTS
The author(s) declare that there is no conflict of interests.