The difference between every two terms of this series constitutes an odd arithmetic series 3, 5, 7, 9 ... The solution is as follows:
a2 - a 1 = 3
a3 - a2 = 5
a4 - a3 = 7
..............
a(n)-a(n- 1)= 2 *(n- 1)+ 1
a(n+ 1)-a(n)=2*n+ 1
Add the two sides of all the formulas to get: a (n+1)-a1= (3+2 * n+1) * n/2 = n * (n+2) = [(n+1)]
So there is a (n) general term: a (n)-0 = (n-1) (n+1) = n 2-1.
a(n)=n^2- 1，
n= 100，
a( 100)= 100^2- 1 = 9999