  In this paper, we are going to perform the shuffle products of $Z_-(n) =
\sum_{a+b=m} (-1)^{b} \zeta(\{1\}^{a},b+2)$ and $Z_+^\star(n) = \sum_{c+d=n}
\zeta^{\star}(\{1\}^{c},d+2)$ with $m+n = p$. The resulted shuffle relation is
a weighted sum formula given by \begin{equation*}
  \frac{(p+1)(p+2)}{2} \zeta(p+4)
  =\sum_{m+n=p} \sum_{|\boldsymbol{\alpha}|=p+3}
  \zeta(\alpha_{0}, \alpha_{1}, \ldots, \alpha_{m}, \alpha_{m+1}+1)
  \sum_{a+b+c=m}
  \Bigl( W_{\boldsymbol\alpha}(a,b,c) + W_{\boldsymbol\alpha}(a,b,c=0)
  + W_{\boldsymbol\alpha}(a=0,b,c) + W_{\boldsymbol\alpha}(a=0,b=m,c=0) \Bigr),
\end{equation*} where $W_{\boldsymbol\alpha}(a,b,c) =
2^{\sigma(a+b+1)-\sigma(a)-(b+1)} (1-2^{1-\alpha_{a+b+1}}\ \ )$, with
$\sigma(r) = \sum_{j=0}^{r} \alpha_{j}$.
