I am trying to work through this identity on page 151 of Paulo Ribenboim's "13 Lectures on Fermat's Last Theorem" $\sum_{j=1}^n (-1)^{j-1} j^{2k+1} = (-1)^{n+1}\{ \frac{n^{2k+1}}{2} + \binom{2k+1}{1}\frac{2^2-1}{2}B_{2}n^{2k} +\binom{2k+1}{3}\frac{2^4-1}{4}B_{4}n^{2k-2}+ \dots+ \binom{2k+1}{2r-1}\frac{2^{2r}-1}{2r}B_{2r}n^{2k+2-2r} + \dots +\frac{2^{2k+2}-1}{2k+2}B_{2k+2} \} +\frac{2^{2k+2}-1}{2k+2}B_{2k+2}$
I have attempted to use the formulae for Euler and Bernoulli polynomials found here: https://dlmf.nist.gov/24 and attempted to use a formula for alternating sum of powers of consecutive integers found here https://arxiv.org/pdf/math/0508233.pdf together with the formulae previously mentioned but I either have powers of $n$ to odd powers that I cant get rid of, or I find myself getting lost in notation and unable to simplify. Is this the correct method to solve this problem? Or should I be attacking it in a different way, if so how? Thanks in advance.
Hmm, don't know whether this is something helpful, I'm a bit dense at the moment.
What I recognize is a difficult reformulation of the Faulhaber-polynomials using the Euler's $ \zeta() \to \eta() $ - conversion. I've not yet(?) checked all the documented conversion-coefficients... ;-) but I think that in general you've taken this correctly.
Here is how I've done the alternating sum of like powers using the inverse of the (Pascal-matrix + I), which lead implicitely to your complicated modified Faulhaber/Bernoulli-coefficicients: it seems much simpler and better to understand.
This uses initially $$ H = (P + I)^{-1} = \small \begin{bmatrix} 1/2 & . & . & . & . & . & . & . \\ -1/4 & 1/2 & . & . & . & . & . & . \\ 0 & -1/2 & 1/2 & . & . & . & . & . \\ 1/8 & 0 & -3/4 & 1/2 & . & . & . & . \\ 0 & 1/2 & 0 & -1 & 1/2 & . & . & . \\ -1/4 & 0 & 5/4 & 0 & -5/4 & 1/2 & . & . \\ 0 & -3/2 & 0 & 5/2 & 0 & -3/2 & 1/2 & . \\ 17/16 & 0 & -21/4 & 0 & 35/8 & 0 & -7/4 & 1/2 \end{bmatrix} \\ $$ With this we introduce something I call "Vandermonde-vectors", simply a vector of consecutive powers of its argument, say $V(x)= [1,x,x^2,x^3,...,x^{n-1}]$ up to some chosen dimension $n$.
If we postmultiply $H \cdot V(1)$ to create polynomials from the row-coefficients, then we get in each row $r$ the infinite alternating sum $\eta(-r,1) = 1^r - 2^r - 3^r + \cdots - \cdots $ or in general $$H \cdot V(x) = E(x) \\ \small \text{ where } E=\operatorname{Column}[\eta(0,x),\eta(-1,x),\eta(-2,x),...,\eta(1-n,x)] $$ Of course, if we use two $V(x)$ -vectors we can compute the finite sums, let's say $$ \small \begin{array} {lll} H \cdot (V(1) + V(2)) = \operatorname{Col}[1,1,1,1...,1]\\ H \cdot (V(1) - V(3)) = \operatorname{Col}[1-1,1-2,1-2^2,1-2^3,...,1-2^{n-1}] \\ H \cdot (V(1) + V(4)) = \operatorname{Col}[1-1+1,1-2+3,1-2^2+3^2,1-2^3+3^3,...,1-2^{n-1}+3^{n-1}] \\ \end{array}$$ I've put the examples together: $$ H \cdot \small{ \operatorname{Concat}[V(1), V(1)+V(2) , V(1)-V(3), V(1)+V(4), V(1)-V(5)]} \\ =\small \begin{bmatrix} 1/2 & 1 & 0 & 1 & 0 \\ 1/4 & 1 & -1 & 2 & -2 \\ 0 & 1 & -3 & 6 & -10 \\ -1/8 & 1 & -7 & 20 & -44 \\ 0 & 1 & -15 & 66 & -190 \\ 1/4 & 1 & -31 & 212 & -812 \\ 0 & 1 & -63 & 666 & -3430 \\ -17/16 & 1 & -127 & 2060 & -14324 \end{bmatrix} $$ If you need only the odd powers in the exponents, take just the results of odd row indexes.
This matrixmultiplication is in the spirit of the Faulhaber-polynomials only that we find different coefficients, not the Bernoulli-numbers/polynomials (which are hidden $\zeta()$-values at negative arguments) but numbers, which one could call $\eta$-numbers (perhaps they have already a name, don't know).
If you need only the solutions, this might be enough, if you need the derivation, it is not difficult, and I have it in some small treatize see treatize at web-index scroll down to "Summing of like powers - solved using the Pascalmatrix...," . The treatize not so much a compact mathematical article, but more-or-less the protocol of my naîve exploration of how I found out about this nice problems of alternating and non-alternating sums of like powers.
update: Upps - I shouldn't forget to remark that this is all explored, calculated and documented using the software Pari/GP which supplies amazing matrix-calculation functions. I seem to always forget to mention this option because it is so easy usable that it is for me like using a pocket-calculator, not worth a specific mention. But if you don't have it, Pari/GP can be downloaded free for personal use from the Pari-group at university Bordeaux.
Over the years I've come across some more learned articles on the Pascal-matrix, but one is perhaps the most interesting for your case:
*Sheng-Liang Yang and Zhong-Kui Liu: "Explicit inverse of the Pascal Matrix plus One"
Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.
This paper presents a simple approach to invert the matrix $P_n + I_n$ by applying the Euler polynomials and Bernoulli numbers, where $P_n$ is the Pascal matrix.