n & gt=2

An = sn-sn-1= (an 2-(an-1) 2)/4+(an-an-1)/2

Square difference expands the first item on the right, and then sorts out the available items.

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

Because the positive term sequence an+an- 1 is not equal to 0, so

An-an- 1 = 2, indicating that the series is arithmetic progression.

By a1= a12/4+a1/2-3/4, and a1>; 0 get

a 1 = 3

So an = 3+2(n- 1) = 2n+ 1.

(2)

According to 1, BN = 2n/3 (n- 1) is obtained.

b 1 = 2

n & gt=2

BN =1/3 * 2 (n-1)/3 (n-2)+2/3 (n-1), that is,

bn = 1/3 * bn- 1+2/3^(n- 1)

bn- 1 = 1/3 * bn-2+2/3^(n-2)

bn-2 = 1/3 * bn-3 + 2/3^(n-3)

......

B2 = 1/3 * b 1+2/3^ 1

Add the above formula.

sn-b 1 = 1/3 sn- 1+ 1-( 1/3)^(n- 1)

sn = 1/3 * sn- 1+3- 1/3^(n- 1)

The following formula for finding the general term of Sn has not found a suitable deformation method. . .