已知正向等差数列an中,其前n项和为sn,满足2sn=anan+1,求数列an的通项公式,设bn=sn

已知正向等差数列an中,其前n项和为sn,满足2sn=anan+1,求数列an的通项公式,设bn=sn

第二问怎么写,求详细解法.

a(n) = a + (n-1)d,a>0,d>0.
s(n) = na + n(n-1)d/2.
2s(1) = 2a(1) = 2a = a(1)a(n+1) = a(a+d),
0 = a(a+d) - 2a = a(a+d-2).
a + d = 2.
2s(2) = 2(2a + d) = a(2)a(3) = (a+d)(a+2d) = 2a + 2(a+d) = (a+d)[2(a+d)-a] = 2(a+d)^2 - a(a+d),
0 = 2(a+d)^2 - (a+2)(a+d) - 2a = 8 - 2(a+2) - 2a = 8 - 2a - 4 - 2a = 4 - 4a = 4(1-a),a = 1.
d = 2-a = 1.
a(n) = 1 + (n-1) = n.
s(n) = n(n+1)/2.
b(n) = [s(n)-1]/2^[a(n)] = [n(n+1)/2-1]/2^n = n(n+1)/2^(n+1) - 1/2^n,
t(n) = b(1)+b(2)+b(3) + ...+ b(n-1)+b(n)
= 1*2/2^2 + 2*3/2^3 + 3*4/2^4 + ...+ (n-1)n/2^n + n(n+1)/2^(n+1) - 1/2-1/2^2 - 1/2^3 - ...- 1/2^(n-1) - 1/2^n
= c(n) - 1/2[1 + 1/2 + ...+ (1/2)^(n-1)]
= c(n) - 1/2[1 - (1/2)^n]/(1-1/2)
= c(n) - [1 - 1/2^n]
= c(n) - 1 + 1/2^n.
c(n) = 1*2/2^2 + 2*3/2^3 + 3*4/2^4 + ...+ (n-1)n/2^n + n(n+1)/2^(n+1),
2c(n) = 1*2/2 + 2*3/2^2 + 3*4/2^3 + ...+ (n-1)n/2^(n-1) + n(n+1)/2^n,
c(n) = 2c(n) - c(n) = 2*1/2 +2*2/2^2 + 2*3/2^3 + ...+ 2*n/2^n - n(n+1)/2^(n+1)
= 1 + 2/2 + 3/2^2 + ...+ n/2^(n-1) - n(n+1)/2^(n+1)
= d(n) - n(n+1)/2^(n+1).
d(n) = 1 + 2/2 + 3/2^2 + 4/2^3+...+(n-1)/2^(n-2) + n/2^(n-1),
2d(n) = 2 + 2 + 3/2 + 4/2^2 + ...+(n-1)/2^(n-3) + n/2^(n-2),
d(n) = 2d(n) - d(n) = 2 + 1 + 1/2 + 1/2^2 + ...+ 1/2^(n-2) - n/2^(n-1)
= 2 + [1 - 1/2^(n-1)]/(1-1/2) - n/2^(n-1)
= 2 + 2[1 - 1/2^(n-1)] - n/2^(n-1)
= 2 + 2 - (n+2)/2^(n-1)
= 4 - (n+2)/2^(n-1).
c(n) = d(n) - n(n+1)/2^(n+1)
= 4 - (n+2)/2^(n-1) - n(n+1)/2^(n+1).
t(n) = c(n) - 1 + 1/2^n
= 4 - (n+2)/2^(n-1) - n(n+1)/2^(n+1) - 1 + 1/2^n
= 3 -[n(n+1) + 4(n+2) - 2]/2^(n+1)
= 3 - [n^2 + n + 4n + 8 - 2]/2^(n+1)
= 3 - [n^2 + 5n + 6]/2^(n+1)
= 3-(n+2)(n+3)/2^(n+1)
< 3