(20 14? The positive term sequence {an} is known, and its first n terms and Sn satisfy 8Sn=an2+4. ...

Solution: Solution: (i) From 8Sn=an2+4an+3 ①.

Get 8sn-1= an-12+4an-1+3 (n ≥ 2, n∈N) ②.

①-②De:8an =(an-an- 1)(an+an- 1)+4an-4an- 1，

Ordered: (an-an-1-4) (an+an-1) = 0 (n ≥ 2, n∈N),

∫{ an} is a positive sequence,

∴ an+an- 1 > 0, then an-an- 1=4(n≥2, n∈N).

∴{an} is the arithmetic progression with an error of 4,

From 8a1= a12+4a1+3, we get a 1=3 or a 1= 1.

When a 1=3, a2=7, and a7=27, the requirement that a2 is the median ratio of a 1 and a7 is not satisfied.

When a 1= 1, a2=5, a7=25, so a2 is the proportional median of A1and a7.

∴an= 1+(n- 1)×4=4n-3；

(ii) from an=4n-3, bn=[log2 (

An +3

four

)]=[log2n]，

The symbol [x] represents the largest integer not exceeding the real number X. When 2m ≤ n < 2m+ 1, [log2n]=m,

Let s = b1+B2+B3+… b2n = [log21]+[log22]+… [log22n]

= 0+ 1+ 1+2+…+3+…+4+…+n- 1+…+n

∴s= 1×2 1+2×22+3×23+4×24+(n- 1)×2n- 1+n

①

2S = 1×22+2×23+3×24+4×25+(n- 1)×2n+2n

②

①-② Obtain:

-S = 2+22+23+24+…+2n- 1-(n- 1)2n-n

=

2( 1-2n- 1)

1-2

-(n- 1)2n-n=(2-n)2n-n-2

∴S=(n-2)2n+n+2，

That's b1+B2+B3+… B2N = (n-2) 2n+n+2.