Given that the sum of the first n terms of a sequence an is Sn, and an = 3 ^ n + 2n, then SN is equal to

Given that the sum of the first n terms of a sequence an is Sn, and an = 3 ^ n + 2n, then SN is equal to

an=3ⁿ+2n
Sn=a1+a2+...+an
=(3+3²+...+3ⁿ)+2(1+2+...+n)
=3×(3ⁿ-1)/(3-1)+2n(n+1)/2
=3^(n+1) /2 -3/2 +n²+n
3 ^ (n + 1) is the power of 3

Let A1 = 1 and (2n + 1) an = (2n-3) a (n-1), (n ≥ 2) in the sequence {an}, find {an}, Sn

Let (2n + 1) an = (2n-3) a (n-1) transform an / a (n-1) = (2n-3) / (2n + 1), then A2 / A1 = 1 / 5, A3 / A2 = 3 / 7, A4 / A3 = 5 / 9. A (n-1) / A (n-2) = (2n-5) / (2n-1), an / a (n-1) = (2n-3) / (2n + 1) multiply each item in turn to get 3 / (2n-1) (2n + 1) = an / A1 = an, so an = 3 / (2n-1) (2n + 1) = (3 /

Given the first n terms of the sequence {an} and the 2n square with Sn equal to 3 times minus n plus one, the general term formula is discussed

Sn=3(2n²－n＋1)
(1) When n = 1, A1 = S1 = 6;
(2) When n ≥ 2, an = SN-S (n-1) = 12n-9
Then:
.{ 6 (n=1)
an={ 12n－9 (n≥2)

It is known that the sum of the first n terms of the sequence an is Sn, Sn is equal to 1 / 3 (an minus 1)

sn=1/3(an-1)