User:Henri10617/sandbox



In mathematics, Sedrakyan's (power sums) triangle, is a self-constructive Pascal-type triangle that expresses the sum of the k-th powers of the first n positive integers


 * $$\sum_{i=1}^n i^k = 1^k + 2^k + 3^k + \cdots + n^k$$

as a (k + 1)th-degree polynomial function of n, it only uses the coefficients of the previous line (see the figure) and coefficients involving Bernoulli numbers Bj. This result was published recently in the book Algebraic Inequalities.

In 1631 Prussian mathematician Johann Faulhaber (1580-1635) published (edition of Academiae Algebrae) the general formula for the sums of powers of the first n positive integers, he knew at least the first 17 cases, as well as the existence of the Faulhaber polynomials for odd powers.

In 1713 Swiss mathematician Jacob Bernoulli published (Summae Potestatum in the book Ars Conjectandi) the general formula for the sums of the k-th powers of the first n positive integers as a (k + 1)th-degree polynomial function of n, the coefficients involving Ars Conjectandis Bj. Nevertheless none of these formulas had a self-constructive property and had a very complicated form. The power sums triangle and its self-constructive property makes this formula extremely easy to use and to remember (even for a high-school student).

Considered “triangle”. The following “triangle” is considered (see the figure), where k stands for the power of the considered sum, $$n$$, $$n^2$$, $$n^3$$, $$n^4$$, $$n^5$$,…, $$n^{k+1}$$ stand for the terms of the (k + 1)th-degree degree polynomial of variable n that the considered sum is equal to, the number written in the intersection of each row and each column stands for the coefficient of the corresponding term of this polynomial.

Self-construction principle. Considering this “triangle” diagonally, we observe that the coefficients on the dashed diagonals remain unchanged, the denominator of each coefficient on the solid line increases by one for each next row (in other words, this coefficient is equal to $$1/(k+1)$$), the arrows indicate how each next row can be recurrently constructed using the previous row (each time one needs to multiply the coefficient of the previous row by a fraction whose nominator is equal to the value of k written in that row and denominator is equal to the power of n written in that column). Note that Bk=0, if k is odd (k>1), so every odd row (starting from the 3rd row starts with 0 and every even row k=2m starts with B2m, where m is a positive integer.

Example (application). Calculate the sum $$1^5 + 2^5 + 3^5 + \cdots + n^5$$.

Using this triangle we obtain that $$1^5 + 2^5 + 3^5 + \cdots + n^5$$ = B4$$\cdot 5/2 n^2$$+B2$$\cdot 3/2 \cdot 4/3 \cdot 5/4 \cdot n^4+1/2 \cdot n^5+1/6 \cdot n^6$$

Thus, it follows that

$$1^5 + 2^5 + 3^5 + \cdots + n^5 = -1/30 \cdot (5n^2)/2 + 1/6 \cdot (5n^4)/2 + n^5/2 + n^6/6.$$

Therefore, we deduce that

$$1^5 + 2^5 + 3^5 + \cdots + n^5 = -n^2/12+(5n^4)/12+n^5/2+n^6/6.$$