The series $\frac1{3^k}+\frac1{6^k}+\frac1{10^k}+\cdots$ for integers $k>1$ at the triangular numbers can be written as $$\sum_{i=2}^\infty\frac1{\left(\frac{i(i+1)}2\right)^k}=2^k\sum_{i=2}^\infty\frac1{(i(i+1))^k}$$ At first glance, the zeta function may be of use. Partial fractions can be used to evaluate the sum, but only for small values of $k$ since the general formula is $$\frac1{i^k(i+1)^k}=\frac{a_1}i+\frac{a_2}{i^2}+\cdots+\frac{a_k}{i^k}+\frac{b_1}{i+1}+\frac{b_2}{(i+1)^2}+\cdots+\frac{b_k}{(i+1)^k}.$$ Of course, once the expression in the sum is presented as such, $\zeta(\cdot)$ can replace each individual term. However, apart from this tedious method, I do not see a way to get rid of the summation.
Is there another way?
Just a few values. $$\left( \begin{array}{cc} 1 & 1 \\ 2 & -13+\frac{4 \pi ^2}{3} \\ 3 & 79-8 \pi ^2 \\ 4 & -561+\frac{160 \pi ^2}{3}+\frac{16 \pi ^4}{45} \\ 5 & 4031-\frac{1120 \pi ^2}{3}-\frac{32 \pi ^4}{9} \\ 6 & -29569+2688 \pi ^2+\frac{448 \pi ^4}{15}+\frac{128 \pi ^6}{945} \\ 7 & 219647-19712 \pi ^2-\frac{3584 \pi ^4}{15}-\frac{256 \pi ^6}{135} \\ 8 & -1647361+146432 \pi ^2+\frac{5632 \pi ^4}{3}+\frac{2048 \pi ^6}{105}+\frac{256 \pi ^8}{4725} \\ 9 & 12446719-1098240 \pi ^2-\frac{73216 \pi ^4}{5}-\frac{11264 \pi ^6}{63}-\frac{512 \pi ^8}{525} \\ 10 & -94595073+\frac{24893440 \pi ^2}{3}+\frac{1025024 \pi ^4}{9}+\frac{292864 \pi ^6}{189}+\frac{11264 \pi ^8}{945}+\frac{2048 \pi ^{10}}{93555} \end{array} \right)$$