Faulhaber's formula

Faulhaber's formula concerns expressing the sum of the pth powers of the first n positive integers \sum_{k=1}^{n} k^p = 1^p + 2^p + 3^p + \cdots + n^p as a (p + 1)th-degree polynomial function of n. The first few examples are well known. For p = 0, we have \sum_{k=1}^n k^0 = \sum_{k=1}^n 1 = n . For p = 1, we have the triangular numbers \sum_{k=1}^n k^1 = \sum_{k=1}^n k = \frac{n(n+1)}{2} = \frac12(n^2 + n) . For p = 2, we have the square pyramidal numbers \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} = \frac13(n^3 + \tfrac32 n^2 + \tfrac12n) . The coefficients of Faulhaber's formula in its general form involve the second Bernoulli numbers B_j^+ which nearly coincide with the first Bernoulli numbers denoted B_j^- (or simply B_j); the sole exception is at j=1, where B_1^-=-\tfrac12 but B_1^+=\tfrac12. The Bernoulli numbers begin \begin{align} B_0 &= 1 & B_1^+ &= \tfrac12 & B_2 &= \tfrac16 & B_3 &= 0 \\ B_4 &= -\tfrac{1}{30} & B_5 &= 0 & B_6 &= \tfrac{1}{42} & B_7 &= 0 , \end{align} Then Faulhaber's formula is that \sum_{k=1}^n k^{p} = \frac{1}{p+1} \sum_{r=0}^p \binom{p+1}{r} B_r^+ n^{p+1-r} . Here, the B_j are the Bernoulli numbers as above, and \binom{p+1}{r} = \frac{(p+1)!}{(p-r+1)!\,r!} = \frac{(p+1)p(p-1) \cdots (p-r+3)(p-r+2)}{r(r-1)(r-2)\cdots2\cdot 1} is the binomial coefficient "p + 1 choose r". ==Examples==

So, for example, one has for p = 4, \begin{align} 1^4 + 2^4 + 3^4 + \cdots + n^4 &= \frac{1}{5} \sum_{j=0}^4 {5 \choose j} B_j^+ n^{5-j}\\ &= \frac{1}{5} \left(B_0 n^5+5B_1^+n^4+10B_2n^3+10B_3n^2+5B_4n\right)\\ &= \frac{1}{5} \left(n^5 + \tfrac{5}{2}n^4+ \tfrac{5}{3}n^3- \tfrac{1}{6}n \right) .\end{align} The first seven examples of Faulhaber's formula are \begin{align} \sum_{k=1}^n k^0 &= \frac{1}{1} \, \big(n \big) \\ \sum_{k=1}^n k^1 &= \frac{1}{2} \, \big(n^2 + \tfrac{2}{2} n \big) \\ \sum_{k=1}^n k^2 &= \frac{1}{3} \, \big(n^3 + \tfrac{3}{2} n^2 + \tfrac{ 3}{6} n \big) \\ \sum_{k=1}^n k^3 &= \frac{1}{4} \, \big(n^4 + \tfrac{4}{2} n^3 + \tfrac{ 6}{6} n^2 + 0n \big) \\ \sum_{k=1}^n k^4 &= \frac{1}{5} \, \big(n^5 + \tfrac{5}{2} n^4 + \tfrac{10}{6} n^3 + 0n^2 - \tfrac{ 5}{30} n \big) \\ \sum_{k=1}^n k^5 &= \frac{1}{6} \, \big(n^6 + \tfrac{6}{2} n^5 + \tfrac{15}{6} n^4 + 0n^3 - \tfrac{15}{30} n^2 + 0n \big) \\ \sum_{k=1}^n k^6 &= \frac{1}{7} \, \big(n^7 + \tfrac{7}{2} n^6 + \tfrac{21}{6} n^5 + 0n^4 - \tfrac{35}{30} n^3 + 0n^2 + \tfrac{7}{42}n \big) . \end{align} == History ==

Ancient period The history of the problem begins in antiquity, its special cases arising as solutions to related inquiries. The case p = 1 coincides historically with the problem of calculating the sum of the first n terms of an arithmetic progression. In chronological order, early discoveries include: : 1+2+\dots+n=\frac{1}{2}n^2+\frac{1}{2}n , a formula known by the Pythagorean school for its connection with triangular numbers. : 1+3+\dots+2n-1=n^2, a result showing that the sum of the first n positive odd numbers is a perfect square. This formula was likely also known to the Pythagoreans, who in constructing figurate numbers realized that the gnomon of the n th perfect square is precisely the n th odd number. :1^2+2^2+\ldots+n^2=\frac{1}{3}n^3+\frac{1}{2}n^2+\frac{1}{6}n, a formula that calculates the sum of the squares of the first n positive integers, as demonstrated in Spirals, a work of Archimedes. Middle period Over time, many other mathematicians became interested in the problem and made various contributions to its solution. These include Aryabhata, Al-Karaji, Ibn al-Haytham, Thomas Harriot, Johann Faulhaber, Pierre de Fermat and Blaise Pascal who recursively solved the problem of the sum of powers of successive integers by considering an identity that allowed to obtain a polynomial of degree m + 1 already knowing the previous ones. , 1713 In 1713, Jacob Bernoulli published under the title Summae Potestatum an expression of the sum of the p powers of the n first integers as a (p + 1)th-degree polynomial function of n, with coefficients involving numbers B_j, now called Bernoulli numbers: : \sum_{k=1}^n k^{p} = \frac{n^{p+1}}{p+1}+\frac{1}{2}n^p+{1 \over p+1} \sum_{j=2}^p {p+1 \choose j} B_j n^{p+1-j}. Introducing also the first two Bernoulli numbers (which Bernoulli did not), the previous formula becomes \sum_{k=1}^n k^p = {1 \over p+1} \sum_{j=0}^p {p+1 \choose j} B_j^+ n^{p+1-j}, using the Bernoulli number of the second kind for which B_1^+=\frac{1}{2}, or \sum_{k=1}^n k^p = {1 \over p+1} \sum_{j=0}^p (-1)^j{p+1 \choose j} B_j n^{p+1-j}, using the Bernoulli number of the first kind B_j=B_j^- for which B_1=-\frac{1}{2}. A rigorous proof of these formulas and Faulhaber's assertion that such formulas would exist for all odd powers took until , two centuries later. Jacobi benefited from the progress of mathematical analysis using the development in infinite series of an exponential function generating Bernoulli numbers. Modern period In 1982, A.W.F. Edwards published an article showing that Pascal's identity can be expressed by means of triangular matrices containing a modified Pascal's triangle: :\begin{pmatrix} n\\ n^2\\ n^3\\ n^4\\ n^5\\ \end{pmatrix}=\begin{pmatrix} 1 &0 &0 &0 &0\\ 1&2&0 &0 &0 \\ 1&3&3&0 &0 \\ 1&4&6&4 &0 \\ 1&5&10&10&5 \end{pmatrix}\begin{pmatrix} n\\ \sum_{k=0}^{n-1} k^1\\ \sum_{k=0}^{n-1} k^2\\ \sum_{k=0}^{n-1} k^3\\ \sum_{k=0}^{n-1} k^4\\ \end{pmatrix} This example is limited by the choice of a fifth-order matrix, but the underlying method is easily extendable to higher orders. Writing the equation as \vec{N}=A\vec{S} and multiplying the two sides of the equation to the left by A^{-1} , we obtain A^{-1}\vec{N}=\vec{S} , thereby arriving at the polynomial coefficients without directly using the Bernoulli numbers. Expanding on Edwards' work, some authors researching the power-sum problem have taken the matrix path, leveraging useful tools such as the Vandermonde vector. Other researchers continue to explore through the traditional analytic route, generalizing the problem of the sum of successive integers to any arithmetic progression.{{cite journal|first=Do|last=Tan Si|title=Obtaining Easily Sums of Powers on Arithmetic Progressions and Properties of Bernoulli Polynomials by Operator Calculus == Polynomials calculating sums of powers of arithmetic progressions ==

Consider the problem of finding polynomials S_{h,d}^p for any nonnegative integer p such that : S_{h,d}^p(n)=\sum_{k=0}^{n-1}(h+kd)^p = h^p+(h+d)^p+\cdots+(h+(n-1)d)^p, given complex numbers h and d \ne 0. Faulhaber's formula handles the simple case S_{1,1}^p(n) = \sum_{k=0}^{n-1}(1+k)^p=1^p+2^p+\dots+n^p. The polynomials S_{1,2}^p(n)=\sum_{k=0}^{n-1}(1+2k)^p=1^p+3^p+\dots+(2n-1)^p compute sums of powers of successive odd numbers, and so on. In general, such polynomials exist for any arithmetic progression. Matrix method The general case is solved using the following matrix formula: :\vec{S_{h,d}}(n)=T(h,d)A^{-1}\vec{N_n}, with row-column definitions :[\vec{S_{h,d}}(n)]_{r}:=S_{h,d}^{r-1}(n),\quad [\vec{N_n}]_r:=n^{r}, :[T(h,d)]_{r,c}:=\begin{cases}0 &\text{if } c>r\\ \binom{r-1}{c-1}h^{r-c}d^{c-1}&\text{if } c\leq r\end{cases}~, \quad [A]_{r,c}:=\begin{cases} 0 &\text{if }c>r\\ \binom{r}{c-1} &\text{if }c\leq r\end{cases}~, \quad where r (row) and c (column) are bound by a given matrix order m. Example To generalize up to p = 4, apply the above formula for matrix order m = 5: : \begin{pmatrix} S_{h,d}^0({n})\\ S_{h,d}^1(n)\\ S_{h,d}^2(n)\\ S_{h,d}^3(n)\\ S_{h,d}^4(n) \end{pmatrix}= \begin{pmatrix} 1& 0& 0& 0&0\\ h& d& 0& 0&0\\ h^2& 2hd& d^2& 0&0\\ h^3& 3h^2d& 3hd^2& d^3&0\\ h^4& 4h^3d& 6h^2d^2& 4hd^3& d^4\\ \end{pmatrix} \begin{pmatrix} 1& 0& 0& 0&0\\ 1& 2& 0& 0&0\\ 1& 3& 3& 0&0\\ 1& 4& 6& 4&0\\ 1& 5& 10& 10&5\\ \end{pmatrix}^{-1} \begin{pmatrix} n\\ n^2\\ n^3\\ n^4\\ n^5 \end{pmatrix} Note that the nonzero elements of T(h,d) follow the binomial theorem, and that A is just Pascal's triangle with each row's last element omitted. Letting h=1, d=2 and computing A^{-1}, we have: T(1,2)=\begin{pmatrix} 1& 0& 0& 0& 0\\ 1& 2& 0& 0& 0 \\ 1& 4& 4& 0& 0\\ 1& 6& 12& 8& 0 \\ 1& 8& 24& 32& 16 \\ \end{pmatrix}, \qquad A^{-1}=\begin{pmatrix} 1 \color{black}&0 &0 &0 &0\\ -\frac{1}{2}\color{black}&\frac{1}{2}&0 &0 &0\\ \frac{1}{6}\color{black}&-\frac{1}{2}&\frac{1}{3} &0 &0\\ 0\color{black}&\frac{1}{4}&-\frac{1}{2}&\frac{1}{4} &0\\ -\frac{1}{30}\color{black}&0&\frac{1}{3}&-\frac{1}{2}&\frac{1}{5} \end{pmatrix} Multiplication therefore yields : \begin{pmatrix} S_{1,2}^0({n})\\ S_{1,2}^1({n})\\ S_{1,2}^2({n})\\ S_{1,2}^3({n})\\ S_{1,2}^4({n}) \end{pmatrix}= \begin{pmatrix} 1 &0 &0 &0 &0\\ 0&1&0 &0 &0 \\ -\frac{1}{3}&0&\frac{4}{3} &0 &0 \\ 0&-1&0&2 &0 \\ \frac{7}{15}&0&-\frac{8}{3}&0&\frac{16}{5} \end{pmatrix} \begin{pmatrix} n\\ n^2\\ n^3\\ n^4\\ n^5\\ \end{pmatrix}=\begin{pmatrix} n\\ n^2\\ -\frac{1}{3}n+\frac{4}{3}n^3\\ -n^2+2n^4\\ \frac{7}{15}n-\frac{8}{3}n^3+\frac{16}{5}n^5\\ \end{pmatrix}. Methods involving Bernoulli polynomials The following formula implicitly solves the problem using Bernoulli polynomials: : S_{h,d}^p(n)=\frac{d^p}{p+1}\left(B_{p+1}\left(n+\frac{h}{d}\right)-B_{p+1}\left(\frac{h}{d}\right)\right) == Faulhaber polynomials ==

The term Faulhaber polynomials is used by some authors to refer to another polynomial sequence related to that given above. Write a = \sum_{k=1}^n k = \frac{n(n+1)}{2} . Faulhaber observed that if p is odd then \sum_{k=1}^n k^p is a polynomial function of a. for For p = 1, it is clear that \sum_{k=1}^n k^1 = \sum_{k=1}^n k = \frac{n(n+1)}{2} = a. For p = 3, the result that \sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4} = a^2 is known as Nicomachus's theorem. Further, we have \begin{align} \sum_{k=1}^n k^5 &= \frac{4a^3 - a^2}{3} \\ \sum_{k=1}^n k^7 &= \frac{6a^4 -4a^3 + a^2}{3} \\ \sum_{k=1}^n k^9 &= \frac{16a^5 - 20a^4 +12a^3 - 3a^2}{5} \\ \sum_{k=1}^n k^{11} &= \frac{16a^6 - 32a^5 + 34a^4 - 20a^3 + 5a^2}{3} \end{align} (see , , , , ). More generally, \sum_{k=1}^n k^{2m+1} = \frac{1}{2^{2m+2}(2m+2)} \sum_{q=0}^m \binom{2m+2}{2q} (2-2^{2q})~ B_{2q} ~\left[(8a+1)^{m+1-q}-1\right]. Some authors call the polynomials in a on the right-hand sides of these identities Faulhaber polynomials. These polynomials are divisible by a^2 because the Bernoulli number B_j is 0 for odd j > 1. Inversely, writing for simplicity S_m: = \sum_{k=1}^n k^m, we have \begin{align} 4a^3 &= 3S_5 + S_3 \\ 8a^4 &= 4S_7 + 4S_5 \\ 16a^5 &= 5S_9 + 10S_7 + S_5 \end{align} and generally 2^{m-1} a^m = \sum_{j>0} \binom{m}{2j-1} S_{2m-2j+1}. Faulhaber also knew that if a sum for an odd power is given by \sum_{k=1}^n k^{2m+1} = c_1 a^2 + c_2 a^3 + \cdots + c_m a^{m+1} then the sum for the even power just below is given by \sum_{k=1}^n k^{2m} = \frac{n+\frac12}{2m+1}(2 c_1 a + 3 c_2 a^2+\cdots + (m+1) c_m a^m). Note that the polynomial in parentheses is the derivative of the polynomial above with respect to a. Since a = n(n + 1)/2, these formulae show that for an odd power (greater than 1), the sum is a polynomial in n having factors n^2 and (n + 1)^2, while for an even power the polynomial has factors n, n + 1/2 and n + 1. == Expressing products of power sums as linear combinations of power sums ==

Products of two (and thus by iteration, several) power sums S_m:=\sum_{k=1}^n k^m can be written as linear combinations of power sums with either all degrees even or all degrees odd, depending on the total degree of the product as a polynomial in n , e.g. 30 S_2S_4=-S_3+15S_5+16S_7 . The sums of coefficients on both sides must be equal, which follows by letting n = 1. Some general formulae include: \begin{align} (m+1)S_m^{\;2} &= 2\sum_{j=0}^{\lfloor\frac{m}2\rfloor}\binom{m+1}{2j}(2m+1-2j)B_{2j}S_{2m+1-2j}.\\ m(m+1)S_mS_{m-1}&=m(m+1)B_mS_m+\sum_{j=0}^{\lfloor\frac{m-1}2\rfloor}\binom{m+1}{2j}(2m+1-2j)B_{2j}S_{2m-2j}.\\ 2^{m-1} S_1^{\;m} &= \sum_{j=1}^{\lfloor\frac{m+1}2\rfloor} \binom{m}{2j-1} S_{2m+1-2j}.\end{align} The latter formula may be used to recursively compute Faulhaber polynomials. Note that in the second formula, for even m the term corresponding to j=\dfrac m2 is different from the other terms in the sum, while for odd m , this additional term vanishes because of B_m=0 . Beardon has published formulas for powers of S_m, including a 1996 paper which demonstrated that integer powers of S_1 can be written as a linear sum of terms in the sequence S_3,\; S_5,\; S_7,\;...: :S_1^{\;N} = \frac{1}{2^N}\sum_{k=0}^{N} {N \choose k}S_{N+k}\left(1-(-1)^{N+k}\right) The first few resulting identities are then :S_1^{\;2} = S_3 :S_1^{\;3} = \frac{1}{4}S_3 + \frac{3}{4}S_5 :S_1^{\;4} = \frac{1}{2}S_5 + \frac{1}{2}S_7 . Although other specific cases of S_m^{\;N} – including S_2^{\;2} = \frac{1}{3}S_3 + \frac{2}{3}S_5 and S_2^{\;3} = \frac{1}{12}S_4 + \frac{7}{12}S_6+ \frac{1}{3}S_8 – are known, no explicit formula for S_m^{\;N} for positive integers m and N has yet been reported. A 2019 paper by Derby proved that: :S_m^{\;N} = \sum_{k=1}^{N}(-1)^{k-1} {N \choose k}\sum_{r=1}^{n}r^{mk}S_m^{\;\;N-k}(r). The m = 1 case replicates Beardon's formula for S_1^{\;N} and confirms the above-stated results for m = 2 and N = 2 or 3. Results for higher powers include: :S_2^{\;4} = \frac{1}{54}S_5 + \frac{5}{18}S_7 + \frac{5}{9}S_9 + \frac{4}{27}S_{11} :S_6^{\;3} = \frac{1}{588}S_8 - \frac{1}{42}S_{10} + \frac{13}{84}S_{12} - \frac{47}{98}S_{14} + \frac{17}{28}S_{16} + \frac{19}{28}S_{18} + \frac{3}{49}S_{20} :S_7^{\;3} = \frac{1}{48}S_{11} - \frac{7}{48}S_{13} + \frac{35}{64}S_{15} - \frac{23}{24}S_{17} + \frac{77}{96}S_{19} + \frac{11}{16}S_{21} + \frac{3}{64}S_{23}. Further generalization is possible by considering the arbitrary power-sum product P := S_{m_1}S_{m_2}S_{m_3} \cdots S_{m_r} given positive integers m_1, \ldots, m_r. For convenience define q := \sum_{k=1}^r{m_k}, and let p_0, \ldots, p_{q+r} be the Maclaurin coefficients of P — i.e. P(n) = \sum_{k=0}^{q+r}{p_kn^k}. It can be shown that :\prod_{k=1}^r{S_{m_k}} = \sum_{k=1}^{q+r-1}{\left(1-(-1)^{q+r+k}\right)p_kS_k}. In particular, the product S_1^{\;r} has trivially retrievable coefficients: :S_1^{\;r}(n) = \left(\frac{n(n+1)}{2}\right)^r = \sum_{k=r}^{2r}{\frac{1}{2^r}\binom{r}{k-r}n^k}. Combining with the above gives :S_1^{\;r} = \frac{1}{2^r}\sum_{k=r}^{2r-1}{\left(1-(-1)^k\right)\binom{r}{k-r}S_k} which is an indexical restatement of Beardon's formula. More generally, :S_m^{\;r} = \sum_{k=1}^{r(m+1)-1}{\left(1-(-1)^{r(m+1)+k}\right)p_kS_k}. Compared to Derby's approach, this formula only requires knowledge of the coefficients of S_m^{\;r}. ==Variations==

• Replacing k with p-k, we find the alternative expression: \sum_{k=1}^n k^p= \sum_{k=0}^p \frac1{k+1}{p \choose k} B_{p-k} n^{k+1}. • Subtracting n^p from both sides of the original formula and incrementing n by 1, we get \begin{align} \sum_{k=1}^n k^{p} &= \frac{1}{p+1} \sum_{k=0}^p \binom{p+1}{k} (-1)^kB_k (n+1)^{p-k+1} \\ &= \sum_{k=0}^p \frac{1}{k+1} \binom{p}{k} (-1)^{p-k}B_{p-k} (n+1)^{k+1}, \end{align} : where (-1)^kB_k = B^-_k can be interpreted as "negative" Bernoulli numbers with B^-_1=-\tfrac12. • We may also expand G(z,n) in terms of the Bernoulli polynomials to find \begin{align} G(z,n) &= \frac{e^{(n+1)z}}{e^z -1} - \frac{e^z}{e^z -1}\\ &= \sum_{j=0}^{\infty} \left(B_j(n+1)-(-1)^j B_j\right) \frac{z^{j-1}}{j!}, \end{align} which implies \sum_{k=1}^n k^p = \frac{1}{p+1} \left(B_{p+1}(n+1)-(-1)^{p+1}B_{p+1}\right) = \frac{1}{p+1}\left(B_{p+1}(n+1)-B_{p+1}(1) \right). Since B_n = 0 whenever n > 1 is odd, the factor (-1)^{p+1} may be removed when p > 0. • It can also be expressed in terms of Stirling numbers of the second kind and falling factorials as \sum_{k=0}^n k^p = \sum_{k=0}^p \left\{{p\atop k}\right\}\frac{(n+1)_{k+1}}{k+1}, \sum_{k=1}^n k^p = \sum_{k=1}^{p+1} \left\{{p+1\atop k}\right\}\frac{(n)_k}{k}. This is due to the definition of the Stirling numbers of the second kind as monomials in terms of falling factorials, and the behaviour of falling factorials under the indefinite sum. :Interpreting the Stirling numbers of the second kind, \left\{{p+1\atop k}\right\}, as the number of set partitions of \lbrack p+1\rbrack into k parts, the identity has a direct combinatorial proof since both sides count the number of functions f:\lbrack p+1\rbrack \to \lbrack n\rbrack with f(1) maximal. The index of summation on the left hand side represents k=f(1), while the index on the right hand side represents the number of elements in the image of f. • There is also a similar (but somehow simpler) expression: using the idea of telescoping and the binomial theorem, one gets ''Pascal's identity'': \begin{align}(n+1)^{k+1}-1 &= \sum_{m = 1}^n \left((m+1)^{k+1} - m^{k+1}\right)\\ &= \sum_{p = 0}^k \binom{k+1}{p} (1^p+2^p+ \dots + n^p).\end{align} :This in particular yields the examples below – e.g., take to get the first example. In a similar fashion we also find \begin{align}n^{k+1} = \sum_{m = 1}^n \left(m^{k+1} - (m-1)^{k+1}\right) = \sum_{p = 0}^k (-1)^{k+p}\binom{k+1}{p} (1^p+2^p+ \dots + n^p).\end{align} • A generalized expression involving the Eulerian numbers A_n(x) is :\sum_{n=1}^\infty n^k x^n=\frac{x}{(1-x)^{k+1}}A_k(x). • Faulhaber's formula was generalized by Guo and Zeng to a -analog. ==Relationship to Riemann zeta function==

Using B_k=-k\zeta(1-k), one can write \sum\limits_{k=1}^n k^p = \frac{n^{p+1}}{p+1} - \sum\limits_{j=0}^{p-1}{p \choose j}\zeta(-j)n^{p-j}. If we consider the generating function G(z,n) in the large n limit for \Re (z), then we find \lim_{n\rightarrow \infty}G(z,n) = \frac{1}{e^{-z}-1}=\sum_{j=0}^{\infty} (-1)^{j-1}B_j \frac{z^{j-1}}{j!} Heuristically, this suggests that \sum_{k=1}^{\infty} k^p=\frac{(-1)^{p} B_{p+1}}{p+1}. This result agrees with the value of the Riemann zeta function \zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s} for negative integers s=-p on appropriately analytically continuing \zeta(s). Faulhaber's formula can be written in terms of the Hurwitz zeta function: \sum\limits_{k=1}^n k^p = \zeta(-p) - \zeta(-p, n+1) ==Umbral form==

In the umbral calculus, one treats the Bernoulli numbers B^0 = 1, B^1 = \frac{1}{2}, B^2 = \frac{1}{6}, ... as if the index j in B^j were actually an exponent, and so as if the Bernoulli numbers were powers of some object B. Using this notation, Faulhaber's formula can be written as \sum_{k=1}^n k^p = \frac{1}{p+1} \big( (B+n)^{p+1} - B^{p+1} \big) . Here, the expression on the right must be understood by expanding out to get terms B^j that can then be interpreted as the Bernoulli numbers. Specifically, using the binomial theorem, we get \begin{align} \frac{1}{p+1} \big( (B+n)^{p+1} - B^{p+1} \big) &= {1 \over p+1} \left( \sum_{k=0}^{p+1} \binom{p+1}{k} B^k n^{p+1-k} - B^{p+1} \right) \\ &= {1 \over p+1} \sum_{k=0}^{p} \binom{p+1}{j} B^k n^{p+1-k} . \end{align} A derivation of Faulhaber's formula using the umbral form is available in The Book of Numbers by John Horton Conway and Richard K. Guy. Classically, this umbral form was considered as a notational convenience. In the modern umbral calculus, on the other hand, this is given a formal mathematical underpinning. One considers the linear functional T on the vector space of polynomials in a variable b given by T(b^j) = B_j. Then one can say \begin{align} \sum_{k=1}^n k^p &= {1 \over p+1} \sum_{j=0}^p {p+1 \choose j} B_j n^{p+1-j} \\ &= {1 \over p+1} \sum_{j=0}^p {p+1 \choose j} T(b^j) n^{p+1-j} \\ &= {1 \over p+1} T\left(\sum_{j=0}^p {p+1 \choose j} b^j n^{p+1-j} \right) \\ &= T\left({(b+n)^{p+1} - b^{p+1} \over p+1}\right). \end{align} ==Notes==