College Algebra
For the general term of the sequence given, write its first four terms: an=(2n+1)!3n2a_{n}=\frac{\left(2n+1\right)!}{3n^2}an=3n2(2n+1)!
2, 10, 560, 7560
10, 50, 5603\frac{560}{3}3560, 7560
2, 10, 5603\frac{560}{3}3560, 6570
2, 10, 5603\frac{560}{3}3560, 7560