Let the first term A1 of the arithmetic sequence {an} be a, and the sum of the first n terms be SN. If s1s2s3 is a proportional sequence, find the general term formula of the sequence {an} …

Let the first term A1 of the arithmetic sequence {an} be a, and the sum of the first n terms be SN. If s1s2s3 is a proportional sequence, find the general term formula of the sequence {an} …


s1=a
s2=2a+d
s3=3a+3d
s2^2=s1*s3
(2a+d)^2=a*(3a+3d)
4a^2+4ad+d^2=3a^2+3ad
a^2+ad+d^2=0
Then solve the equation to get D



If the sum of the first n terms of the sequence an is Sn = -- N ^ 2, find the general term formula of the sequence, and judge whether the sequence is equal difference sequence or equal ratio sequence, please give the proof


Sn=-n^2
a1=S1=-1
When n ≥ 2
an=Sn-S(n-1)=-n^2-[-(n-1)^2]=-n^2+n^2-2n+1=1-2n
Where a1 = - 1 = 1-2 * 1 also conforms to the general formula
So an = 1-2n is an arithmetic sequence



The sum of the first n terms of the sequence (an} is Sn, and satisfies an = - 3Sn * sn-1 (n > = 2), A1 = 1 / 3 1. It is proved that (1 / Sn} is an arithmetic sequence
The sum of the first n terms of the sequence (an} is Sn, and satisfies an = - 3Sn * sn-1 (n > = 2), A1 = 1 / 3
1. Prove that (1 / Sn} is arithmetic sequence
2. Let BN = 2 (1-N) an (n > = 2) prove B2 square + B3 square +. + BN square


1)S[n]-S[n-1] = -3S[n]*S[n-1]
Divide two sides by - s [n] * s [n-1]
1/S[n] -1/S[n-1] =3
2)S[n]=1/(3n)
a[n]= S[n] - S[n-1] =1/( 3(1-n)n )
b[n]=2/(3n)
Because 1 / (n ^ 2) < 1 / ((n-1) n) = 1 / (n-1) - 1 / n
B2 square + B3 square +. + BN square
=4/9*(1/2^2 +1/3^2+...+1/n^2)



It is known that the sum of the first n terms of the sequence {an} is Sn, and A1 = 2, 3Sn = 5an-an-1 + 3sn-1 (n ≥ 2, n ∈ n *) (I) find the general formula of the sequence {an}; (II) let BN = (2n-1) an, find the sum of the first n terms of the sequence {BN}


(I) by 3Sn = 5an-an-1 + 3sn-1  3an = 5an-an-1 (n ≥ 2, n ∈ n *), Anan − 1 = 12, (n ≥ 2, n ∈ n *), so the sequence {an} is an equal ratio sequence with 2 as the first term and 12 as the common ratio, | an = 22-n (Ⅱ) BN = (2n-1) · 22-n  TN = 1 × 2 + 3 × 20 + 5 × 2-1 + + (2n-1) · 22