for-the-general-term-of-the-sequence-given-write-its-first-four-terms-an2n13n2an

Question

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)!​﻿

Accepted Answer

2, 10, ﻿5603\frac{560}{3}3560​﻿, 7560

Answer

2, 10, 560, 7560

Answer

10, 50, ﻿5603\frac{560}{3}3560​﻿, 7560

Answer

2, 10, ﻿5603\frac{560}{3}3560​﻿, 6570

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)!