高三数学真题。

a2 = 2a 1-2+2 = 2a 1 = 2×2 = 4

a3 = 2 a2-3+2 = 2 a2-1 = 2×4-1 = 7

当n≥2时,

an=2a(n-1)-n+2

an-n = 2a(n-1)-2n+2 = 2a(n-1)-2(n-1)= 2[a(n-1)-(n-1)]

(an-n)/[A(n-1)-(n-1)]= 2,为固定值。

a 1-1 = 2-1 = 1,数列{an-n}是以1为第一项,以2为公比的几何级数。

an-n=1×2^(n-1)=2^(n-1)

an=n+2^(n-1)

bn=an/2^(n-1)=[n+2^(n-1)]/2^(n-1)=1+ n/2^(n-1)

Sn=b1+b2+...+bn = 1+1/1+1+2/2+...+1+n/2^(n-1)=n+ 1/1+2/2...+n/2^(n-1)

使cn = 1/1+2/2+...+n/2 (n-1)

则(1/2) cn = 1/2+2+...+(n-1)/2 (n-1)+n/2?

Cn-(1/2)Cn =(1/2)Cn = 1+1/2+...+1/2^(n-1)-n/2?

=1×[1-(1/2)?]/(1-1/2)-n/2?

=2- (n+2)/2?

Cn=4-2(n+2)/2?= 4-n/2^(n-1)-1/2^(n-2)

sn = n+cn = n+4-n/2^(n-1)-1/2^(n-2)