In the sequence {an}, if the first n terms and Sn = 3n-2n ^ 2 (n belongs to n *), then an =? Second, what is the size relationship between Sn and Nan?

In the sequence {an}, if the first n terms and Sn = 3n-2n ^ 2 (n belongs to n *), then an =? Second, what is the size relationship between Sn and Nan?

∵Sn=3n-2n^2
∴an=sn-sn-1=3n-2n^2-3（n-1）+2（n-1）^2=-4n+5,
When n = 1, A1 = S1 = 1
∴an=-4n+5,
.
∵sa-nan=3n-2n^2+4n^2-5n=2n（n-1）≥0
∴sa≥an

Given Sn = 3N ^ 2-2n + 5, find an

Sn=3n^2-2n+5
an=sn-s(n-1)
=3n^2-2n+5-[3(n-1)^2-2(n-1)+5]
=3n²-2n+5-[3n²-6n＋3-2n＋2＋5]
=6n-5
(n>1)
a1=3-2+5=6
Namely
an=6n-5 (n>1)
6 ,n=1

Let n be any integer, and try to prove that n (n + 1) (2n + 1) is a multiple of 6

N (n + 1) (2n + 1) / 6 = 1 ^ 2 + 2 ^ 2 +. + n ^ 2 formula method if you don't know the formula, you can also do this because N and (n + 1) are odd even, so n (n + 1) (2n + 1) is always a multiple of 2 if n = 3K 3 can be divisible by n = 3k, so n (n + 1) (2n + 1) is a multiple of 3, and N = 3K + 1 3 can be divisible by 2n + 1 = 6K + 3, so n (n + 1) (2n +