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