for the problem Sum of series : we just have to calculate Sn the sum of consecutive terms of an arithmetic progression  (suite arithmétique)
with Tn = n^2*(n - 1 )^2 ; (^ is power).
Sn = T1 + T2 + .... + Tn .
with some basic knowledge in math we can prove that Tn = 2*n - 1 . we know that the sum of consecutive terms of an arithmetic progression is (n + k - 1 ) *( (Nth term + Kth term )/2  ). Nth is the last term. Kth is the first term.
T1 = 2 * 1 - 1 = 1 ;
so  Sn = (n + 1 - 1 ) * ( ( 2 * n - 1 + 1 ) / 2 ) = n * (( 2 * n ) / 2 ) ) = n ^ 2 ;
so the solution is n^2 .
then after reading the input n ( n <= 10^10 ) we have to do n % 10^9 + 7 , so we can avoid an overflow ( surpassing the max value of long int (10^18) ) .
We can prove it also by writing some examples on paper :
if n == 1 sol is 1
if n == 2 sol is 4
if n == 3 sol is 9
and so on ...
Sorry for  for the bad explanation for the problem earlier. it happens sometimes. hope that you can understand it now.