(2)an=sn-s(n- 1)=3^n+ 1-[3^(n- 1)+ 1]

=3^n-3^(n- 1)=(3- 1)*3^(n- 1)=2*3^(n- 1)。

(3) Series is not geometric series.

Because an = 2 * 3 (n- 1) and a (n- 1) = 2 * 3 (n-2),

an/2(n- 1)=3，(n & gt=2)

q=3 .

But a 1=4, a2=6, a2/a 1=3/2 is not equal to 3.

So the sequence {{an}} is a geometric series with a common ratio of 3 from the second term except the first term.

But the series {{an}} is not a geometric series.

2. Connect two diagonals of plane A1b1c1d1,B 1D 1, and the intersection point is o,

Then connect AO, DO, the vertical line with intersection a as DO, and the intersection h.

Branch 1: ob1= od1= OA1= oc1= √ 2/2, AO=DO=√6/2.

In the triangle ADO, AO=DO=√6/2, AD= 1,

Using cosine theorem, we know that cosAOD=2/3, so sinAOD=√5/3.

So AH=√30/6.

Because AH is perpendicular to OD, OD is in plane A 1C 1D,

So the length of AH is the distance from a to plane A 1C 1D, which is √30/6.

Third, there are five eligible straight lines.

The right focal coordinate of hyperbola x 2-y 2/2 = 1 is f1(√ 5,0).

The straight line l passing through F 1 passes through the hyperbola at two points a and b,

When the slope of the straight line L does not exist, the straight line L is perpendicular to the X axis, and the equation is: x=√5,

When x=√5, substitute x 2-y 2/2 = 1, and the coordinates of a and b are (√5, -2), (√5, 2).

|AB|=4, so x=√5 is a straight line that meets the conditions.

When the slope of the straight line l exists,

Let the slope of the straight line L be k, then the equation of the straight line L is: y=kx-√5k.

Substituting x 2-y 2/2 = 1, we get: (2-k 2) x 2+2 √ 5k 2 * x-5k 2-2 = 0,

From Vieta's theorem, I get:

x 1+x2=2√5k^2/(k^2-2)、x 1x2=(5k^2+2)/(k^2-2)。

So | x1-x2 | 2 = (x1+x2) 2-4x1x2 = 4 (3k2+4)/(k2-2) 2,

So | x1-x2 | = 2 √ (3k2+4)/| k2-2 |.

So | AB | = √ (K2+1) * | x1-x2 | = 2 √ (K2+1) (3k2+4)/| K2-2 | = 4,

That is, (k 2+ 1) (3k 2+4) = 4 (k 2-2) 2,

Simplification: k 4-23k 2+ 12 = 0.

Because 23 2-4 *12 = 481>; 0 and √ 48 1

So the equation has four different solutions, that is, k has four different values.

Therefore, when the slope of the straight line L exists, four straight lines satisfy the conditions.

To sum up, there are five eligible straight lines.

(3x- 1)/(2-x) < = 1, then:

(3x- 1)/(2-x)- 1 & lt； =0,

(4x-3)/(2-x)& lt； =0,

So 4x-3 =>；; 0 and 2-x

Solution: x>2; Or x < =3/4.

So the solution set of inequality is: {x | x >；; 2 or x