Given that the sequence {an}, Sn is the sum of the first n terms, and satisfies 3an = 2Sn + N, n is a positive integer, it is proved that the sequence {an + 1 / 2} is an equal ratio sequence 1. Given the sequence {an}, Sn is the sum of its first n terms, and satisfies 3an = 2Sn + N, n is a positive integer (1) And prove that the sequence {an + 1 / 2} is equal ratio sequence; (2) Note TN = S1 + S2 + S3 +. + Sn and find the expression of TN 2. It is known that the sequence {an} satisfies A1 = 1. A2 = 2, a (n + 2) = an + a (n + 1) / 2, and N is a positive integer (1) Let BN = a (n + 1) - an and prove that {BN} is an equal ratio sequence (2) Find the general expression of {an} Find the answer to the second question~

Given that the sequence {an}, Sn is the sum of the first n terms, and satisfies 3an = 2Sn + N, n is a positive integer, it is proved that the sequence {an + 1 / 2} is an equal ratio sequence 1. Given the sequence {an}, Sn is the sum of its first n terms, and satisfies 3an = 2Sn + N, n is a positive integer (1) And prove that the sequence {an + 1 / 2} is equal ratio sequence; (2) Note TN = S1 + S2 + S3 +. + Sn and find the expression of TN 2. It is known that the sequence {an} satisfies A1 = 1. A2 = 2, a (n + 2) = an + a (n + 1) / 2, and N is a positive integer (1) Let BN = a (n + 1) - an and prove that {BN} is an equal ratio sequence (2) Find the general expression of {an} Find the answer to the second question~

one
Certificate:
Sn=(3an-n)/2
Sn-1=[3a(n-1)-(n-1)]/2
an=Sn-Sn-1=[3an-3a(n-1)-1]/2
an=3a(n-1)+1
an+1/2=3a(n-1)+3/2=3[a(n-1)+1/2]
(an + 1 / 2) / [a (n-1) + 1 / 2] = 3
{an + 1 / 2} is an equal ratio sequence
Let n = 1
3a1=2a1+1
a1=1
a1+1/2=3/2
Tn=S1+S2+...+Sn
=(1/2)(2S1+2S2+...+2Sn)
=(1/2)(3a1-1+3a2-2+...+3an-n)
=[-n(n+1)/4]+(3/2)(a1+a2+...+an)
=[-n(n+1)/4]+(3/2)[a1+1/2+a2+1/2+...+an+1/2-n/2]
=[-n(n+1)/4]-3n/4+(3/2)(3/2)(3^n-1)/2
=[9(3^n-1)-2n(n+4)]/8