If an = BN, find n

If an = BN, find n

an = 2+3（n-1)=3n -1
bn = -2 +4(n-1)=4n-6
3n-1 =4n-6
n=5

A1 = 1, A2 = 2, when n > 2, Sn = n / 2An + 1, (1) find an (2) find the sum of the first n terms of the sequence {BN} as TN, where BN = 1 / Anan + 1

Sn = n / 2 * a (n + 1) S1 = (1 / 2) A2 = 1 A1 = S1 = 1s (n-1) = (n-1) / 2 * an, so an = SN-S (n-1) = n / 2 * a (n + 1) - (n-1) / 2 * ana (n + 1) / (n + 1) = an / N, so {an / N} is an equal ratio sequence with common ratio 1, A1 / 1 = 1An / N = 1 * 1 ^ (n-1) = 1 (1) an = n (2) BN = 1 / n (n + 1) = 1 / n-1 / (n + 1) TN = (1-1 /

If Sn / TN = 2n / 3N + 1, find an / BN

S(2n-1)=(A1+A(2n-1))×(2n-1)/2
=(A1+A1+(2n-2)d)×(2n-1)/2
=(A1+(n-1)d)×(2n-1)
=An×(2n-1)
In the same way
T(2n-1)=Bn×(2n-1)
[An×(2n-1)]/[Bn×(2n-1)]
=S(2n-1)/T(2n-1)
=2(2n-1)/[3(2n-1)+1]
=(4n-2)/(6n-3+1)
=(2n-1)/(3n-1)
An/Bn=(2n-1)/(3n-1)

Sequence {BN} = 3n-1, the formula for finding the first n terms and Sn of sequence

BN is the arithmetic sequence, B1 = 2, BN = 3n-1. According to the sum formula of arithmetic sequence, Sn = n (2 + 3n-1) / 2