The [tex]n[/tex]-th term is
[tex]U_n = \dfrac14 n^2 (n+1)^2[/tex]
so the 39th term is
[tex]U_{39} = \dfrac14 39^2 40^2 = \boxed{608,400}[/tex]
Observe that
[tex]2^3 + 4^3 + 6^3 = 2^3 + 2^3\times2^3 + 2^3\times3^3 = 8\left(1^3+2^3+3^3)[/tex]
which suggests that
[tex]V_n = 8U_n = \boxed{2n^2(n+1)^2}[/tex]
