For positives integers α1,α2,…,αr\documentclass[12pt]{minimal}
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\begin{document}$$\alpha _{1}, \alpha _{2}, \ldots , \alpha _{r}$$\end{document} with αr≥2\documentclass[12pt]{minimal}
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\begin{document}$$\alpha _{r} \ge 2$$\end{document}, the multiple zeta value or r\documentclass[12pt]{minimal}
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\begin{document}$$r$$\end{document}-fold Euler sum ζ(α1,α2,…,αr)\documentclass[12pt]{minimal}
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\begin{document}$$\zeta (\alpha _{1}, \alpha _{2}, \ldots , \alpha _{r})$$\end{document} is defined by the multiple series ∑1≤n1<n2<⋯<nrn1-α1n2-α2…nr-αr.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \sum _{1 \le n_{1} < n_{2} < \cdots < n_{r}} n_{1}^{-\alpha _{1}} n_{2}^{-\alpha _{2}} \ldots n_{r}^{-\alpha _{r}}. \end{aligned}$$\end{document}In this paper, for integers k,r≥0\documentclass[12pt]{minimal}
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\begin{document}$$k,r\ge 0$$\end{document} and complex numbers μ,λ,\documentclass[12pt]{minimal}
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\begin{document}$$\mu ,\lambda ,$$\end{document} we consider the double weighted sum defined by Ek,r(μ,λ)=∑p+q=kμp∑α=q+r+3ζ(1p,α0,…,αq,αq+1+1)λαq+1\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} E_{k,r}(\mu ,\lambda )=\sum _{p+q=k}\mu ^{p}\sum _{\left| \alpha \right| =q+r+3}\zeta ({\left\{ 1 \right\} ^{p},\alpha _{0},\ldots ,\alpha _{q},\alpha _{q+1}+1})\lambda ^{\alpha _{q+1}} \end{aligned}$$\end{document}and then evaluate Ek,r(2,2),\documentclass[12pt]{minimal}
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\begin{document}$$E_{k,r}(2,2),$$\end{document}Ek,r(2,1),\documentclass[12pt]{minimal}
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\begin{document}$$E_{k,r}(2,1),$$\end{document}Ek,r(1,2),\documentclass[12pt]{minimal}
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\begin{document}$$E_{k,r}(1,2),$$\end{document}Ek,r(1,1)\documentclass[12pt]{minimal}
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\begin{document}$$E_{k,r}(1,1)$$\end{document} and Ek,r(0,1)\documentclass[12pt]{minimal}
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\begin{document}$$E_{k,r}(0,1)$$\end{document} in terms of the special values at positive integers of the Riemann zeta function. Note that Ek,r(0,1)=∑α=k+r+3ζ(α0,…,αk,αk+1+1)\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} E_{k,r}(0,1)=\sum _{\left| \alpha \right| =k+r+3}\zeta (\alpha _{0},\ldots ,\alpha _{k},\alpha _{k+1}+1) \end{aligned}$$\end{document}so our results cover the sum formula ∑α=k+r+3ζ(α0,…,αk,αk+1+1)=ζ(k+r+4)\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \sum _{\left| \alpha \right| =k+r+3}\zeta (\alpha _{0},\ldots ,\alpha _{k},\alpha _{k+1}+1)=\zeta (k+r+4) \end{aligned}$$\end{document}proved by Granville in 1996.