AN= 1/[√n+√(n+ 1)] The denominator is rationalized.

The numerator and denominator are multiplied by [√( n+ 1)-√n] simultaneously.

Simplified as an = √ (n+1)-√ n.

A 1=√2-√ 1

A2=√3-√2

A3=√4-√3

.........

A(n- 1)=√n-√(n- 1)

An=√(n+ 1)-√n

Sum the left and right sides separately.

Sn=√(n+ 1)- 1

Now let Sn=9=√(n+ 1)- 1.

n=99

2.C

Sn=a 1*n+n(n- 1)d/2

pass by

S50=200

S 100-S50=2700

Draw (a conclusion)

50*a 1+ 1225*d=200

100 * a 1+4950 * d = 2700+200

solve

a 1=-20.5

d= 1

3.a:b:c=4: 1:(-2)

According to the meaning of the question, 2b=a+c, let c=kb, then a=(2-k)b(a, C and B are geometric series, so none of them are 0, and k is not equal to 0,2).

And c 2 = BC,

So k 2 = 2-k,

k= 1，-2

A: b: c =1:1:1(shed), or a:b:c=4: 1:(-2).

4.a 1+a3+a5+...+a99=60

a2+a4+a6+a8....+a 100=x

A 1 differs from a2 by d. . . There is also a d between a 100 and a99.

* * * There are 50 d's.

Then A 1+A3+A5+…+A99=x-50d.

S 100=x-50d+x= 145

x=85

So A 1+A3+A5+…+A99=85-25=60.

5.S 13= 156/5

a3+a7+a 10=8，a4+a 1 1=4

a7 = a3+4d a 10 = a3+7d a4 = a3+d a 1 1 = a3+8d

So a3+a7+a10 = a3+a3+4d+a3+7d = 3a3+11d = 8.

a4+a 1 1 = a3+d+a3+8d = 2 a3+9d = 4

3a3+ 1 1d=8，2a3+9d=4

Calculate A3 = 28/5 and D =-4/5.

a 1=a3-2d=36/5

S 13= 156/5