The remainder obtained when $(2 m+1)^{2 n}(m, n \in N)$ is divided by 8 is

$$ \sum_{r=1}^{15} r^2\left(\frac{{ }^{15} C_r}{{ }^{15} C_{r-1}}\right)= $$

$$ \frac{1}{81^n}-{ }^{2 n} C_1 \frac{10}{81^n}+{ }^{2 n} C_2 \frac{10^2}{81^n}-\ldots+\frac{10^{2 n}}{81^n}= $$

If $x$ is positive real number and the first negative term in the expansion of $(1+x)^{\frac{27}{5}}$ is $t_k$, then $k=$