Algebraic manipulation of $(\sum_j \delta_j^2)^4$

Question

I am looking at the solution of a problem from Rene Schilling's Brownian Motion.

Here, I don't understand how to get the second decomposition of the sums. That is, in the second and third group of sums, it seems like the only difference is in the first term, where $\sum\sum\sum\sum_{j<k<l<m}\delta_j \delta_k \delta_l \delta_m$ is replaced by $ \big(\sum_j \delta_j\big)^4$ with the same constant $c_{1,1,1,1}^{'}$ in front, and in the other terms, the structure of the sums are the same except the constants are changed. So there must be some manipulation of constants and sums, however, I can't figure out what it is. I would greatly appreciate any help.

Answer 1

I'll do a two-variable analogue. Suppose we had $$\newcommand{\de}{\delta}a\sum_{j<k}\de_j\de_k+b\sum_j\de_j^2.$$ This equals \begin{align} \frac{a}2\sum_{j\ne k}\de_j\de_k+b\sum_j\de_j^2 &=\frac{a}2\sum_{j,k}\de_j\de_k-\frac a2\sum_j\de_j^2+b\sum_j\de_j^2\\ &=\frac{a}2\left(\sum_{j}\de_j\right)^2+\left(b-\frac a2\right) \sum_j\de_j^2. \end{align}

The four variable version is similar, but with more fiddly details. (I don't blame the author for wanting to suppress them!)