Evaluate: $-T_1^{2k}+T_2^{2k}-T_3^{2k}+T_4^{2k}-T_5^{2k}+T_6^{2k}-\cdots+T_{2n}^{2k}$ $T_n={n(n+1)\over 2}$, observing only the even terms of $T_n$ such as the series below
$$\color{red}{-1^2+3^2=8\times1^2}$$
$$\color{blue}{-1^2+3^2-6^2+10^2=8\times3^2}$$
$$\color{Red}{-1^2+3^2-6^2+10^2-15^2+21^2=8\times6^2}$$
$$\color{blue}{-1^2+3^2-6^2+10^2-15^2+21^2-28^2+36^2=8\times10^2}$$

$$-T_1^2+T_2^2-T_3^2+T_4^2-T_5^2+T_6^2-\cdots+T_{2n}^2$$ 
$$T_2^2-T_1^2+T_4^2-T_3^2+T_6^2-T_5^2-\cdots+T_{2n}^2-T_{2n-1}^2$$ 
$$8(1+2^3+3^3+\cdots n^3)$$
Well-known $1+2^3+3^3+\cdots +n^3=(1+2+3+\cdots n)^2=\left({n(n+1)\over 2}\right)^2=T_n^2$
so we have 
$$8(1+2^3+3^3+\cdots n^3)=8T_n^2$$
My question is: Can we find a general formula for $$-T_1^{2k}+T_2^{2k}-T_3^{2k}+T_4^{2k}-T_5^{2k}+T_6^{2k}-\cdots+T_{2n}^{2k}?$$
$k:=1,2,3,...$

The formula for $k=2$ is
$$
4 n^2 (n + 1)^2 (2 n^4 + 4 n^3 - 2 n + 1)
\\= 16 T_n^2 (8 T_n^2- 4 T_n + 1)
$$
The formula for $k=3$ is
$$
n^2 (n + 1)^2 (32 n^8 + 128 n^7 + 96 n^6 - 160 n^5 - 61 n^4 + 294 n^3 - 33 n^2 - 228 n + 114)
\\= 8 T_n^2 (256 T_n^4 - 384 T_n^3 + 390 T_n^2 - 228 T_n + 57)
$$
Thus, an interesting question is to express the sum as a polynomial in $T_n$, as in Faulhaber polynomials.
