Aman's Greek Letter Challenge

$\lambda=\displaystyle\sum_{n=1}^{\rho} n^{3}\\=\frac{\rho^{2}(1+\rho)^{2}}{4}$

Note that all values of $\lambda$ are perfect squares.

$\zeta=\displaystyle\sum_{\rho=1}^{100} \frac{\rho^{2}(1+\rho)^{2}}{4}\\=0.25(\displaystyle\sum_{1}^{100} \rho^4+2\displaystyle\sum_{1}^{100} \rho^3+\displaystyle\sum_{1}^{100} \rho^2)\\=0.25(2101676680)=525419170$

$\lfloor\frac{-\zeta}{10^7}\rfloor=\lfloor -52.54... \rfloor\\=\boxed{-53}$