Given the sum of the first n terms of the sequence {an} and Sn = - n ^ 2 + 4N 1) find the general term formula of an; 2) find the sum of the first n terms of the sequence {9-2an / 2 ^ n} Because the original problem is the quadratic of negative n, not the quadratic of negative n, so ~ A1 seems to be 3, not 5! The general formula is also wrong!

Given the sum of the first n terms of the sequence {an} and Sn = - n ^ 2 + 4N 1) find the general term formula of an; 2) find the sum of the first n terms of the sequence {9-2an / 2 ^ n} Because the original problem is the quadratic of negative n, not the quadratic of negative n, so ~ A1 seems to be 3, not 5! The general formula is also wrong!

An = SN-S (n-1) = - n ^ 2 + 4N - [- (n-1) ^ 2 + 4 (n-1)] = - n ^ 2 + 4N + (n-1) ^ 2-4 (n-1) = - 2n + 5 (if Sn constant term is 0, A1 need not be verified, otherwise A1) BN = 9-2an / 2 ^ n = 9-2 (- 2n + 5) / 2 ^ n = 9 + (4n-10) / 2 ^ n, let CN = n / 2 ^ n, then SCN = 1 / 2 ^ 1 + 2 / 2 ^ 2 + 3 / 2 ^ 3 + 4 / 2 ^

If the first n terms of the sequence {an} and Sn = N2 + 2n are known, then the general term formula of the sequence {an}=______ ．

∵ Sn = N2 + 2n ①, ∵ sn-1 = (n-1) 2 + 2 (n-1) (n ≥ 2) ②, ① - ②, an = 2n + 1 (n ≥ 2), when n = 1, A1 = S1 = 3, suitable for the above formula, ∵ an = 2n + 1

Given the sequence {an}, {an} = (n + 1) / (N2 (n + 2) 2), then SN=

an=(n+1)/[n²(n+2)²]
=(1/4)(4n+4)/[n²(n+2)²]
=(1/4)[(n+2)²-n²]/[n²(n+2)²]
=(1/4)[1/n²- 1/(n+2)²]
Sn=a1+a2+...+an
=(1/4)[1/1²-1/3²+1/2²-1/4²+...+1/n²-1/(n+2)²]
=(1/4)[(1/1²+1/2²+...+1/n²)-(1/3²+1/4²+...+1/(n+2)²)]
=(1/4)[1/1²+1/2²-1/(n+1)²-1/(n+2)²]
=5/16 -1/[4(n+1)²] -1/[4(n+2)²]

If the first n terms and Sn = n2-1 are known, then the general term an=

Known
S(n)=n²-1
be
S(n-1)=(n-1)²-1=n²-2n-2
a(n)=S(n)-S(n-1)=2n+1