In the series of equal differences an, Sn is the sum of the first n terms, and if Sn 20 then a2 a3 a

sn=na1+(n-1)d;

d=(sn-na1)/(n-1)=(20-na1)/(n-1);

where d is the tolerance, n>1.

a2+a3+a4=3a1+3d=3a1+3(20-na1)/(n-1)=(60-3a1)/(n-1);

If there are no other conditions for the question, this should be the answer.

If it is a series of natural numbers, it can be seen from the question a2+a3+a4: n>=4

sn=na1+(n-1)d=20;

n=(20+d)/(a1+d)>=4;4a1+3d<=20;

and a1 is a natural number, so d=1,4;

When d=1,4a1+3<=20; a1<=17/4;a1<=4;

When a1=1, the first equation gives n=21 2; Abandon it.

When a1=2, the same goes for n=7So a2 + a3 + a4 = 3a1 + 3d = 9

a1=3, the same goes for n=21 4; Abandon it.

a1=4, the same goes for n=21 5, rounded.

When d=4, in the same way, 4a1+3*4<=20; a1<=2;

When a1=1, n+4(n-1)=20, n=24 5, rounded.

When a1=2, 2n+4(n-1)=20, n=4, a2+a3+a4=3a1+3d=18

Here we only talk about the increasing series, and the decreasing ones will not be discussed.

In my opinion, if there are no other conditions, this question is just to let everyone discuss all possible situations, the purpose is to develop the habit of considering the problem holistically, but generally the conditions will be explained in non-discussion situations, such as the above, if it is a series of increasing natural numbers n<7, the answer is determined.

If sn=20, I can only understand that sn is equal to 20, no matter what value n takes is 20, then s1=a1=20

s4=a1+a2+a3+a4=20

Subtract, a2 + a3 + a4 = 0

This question can only be understood in this way, otherwise there will be no answer.

s12=12(a1+a12) 2=21 a1+a12=7 2 a3+a4+a9+a10 =(a3+a10)+(a4+a9) =a1+a12)+(a1+a12) =7 pie Li Fang 2+7 2 =7

Untie the stimulus s4=a1+a2+a3+a4=4a2+2d=22d=(22-4a2) and tolerate the rent of branches 2=3

a1=a2-d=4-3=1

an=a1+nd

n=(an-a1) Changzhao d=9

From a3b3=1 2,s3+s5=21, a1=1, d=1

bn=2/n(n+1)=2(1/n-1/n+1)

Accumulate and prove.

The sum of the first n terms of the equal difference series is sn, bn=1 sn, and a3b3=1 2a3b3=a3 s3=a3 (a1+a2a+a3)=1 2 a1+a2=a3, tolerance d=a1, an=n·a1s5+s3=15a1+6a1=21a1=21 a1=1,an=n,sn=n(n+1) 2,bn=2 n(n+1)the second questiona1=1,a2=2a1 (a1+2)=2 3, a3=2a2 (a2+2)=1 2

1/an=[a(n-1)+2]/2a(n-1)=1/2 + 1/a(n-1)

1 2 + 1 2 + 1 a(n-2) {1 an} into a series of equal differences.

1/an=(1+n)/2,an=2/(1+n)

Let the sum of the first n terms of the difference series be sn, bn=1 sn, and a3b3=1 2, s5+s3=21(1) find bn (2) find the first two terms of bn and tn

b3=1/s3=1/(a1+a2+a3)=1/(3a1+3d)

a3=a1+2d

s3=3a1+3d

s5=5a1+10d

a3b3=1/(3a1+3d)*(a1+2d)=1/2 (1)

s5+s3=(3a1+3d)+(5a1+10d)=21 (2)

From (1) and (2):

a1=d=1 because sn=(a1+an)n 2=(1+n)n 2

1)bn=1/sn=2/[(1+n)n]

2)b1=1,b2=1/3

tn is estimated to be the sum of the first n terms of bn, tn=2 [(1+n)n].

2*[1/1-1/2+1/2-1/3+1/3-1/4...1/(n-1)-1/n]

2*(1-1/n)

2(n-1)/n

Question 2. It is known that the sequence {an} satisfies a1=1, an=2a(n-1) a(n-1)+2

1) Prove that {1 an} is a series of equal differences (2) Find the general term of an.

If the expression of an is incorrect, it should be an=2a(n-1) [a(n-1)+2].

1) an=2a(n-1) [a(n-1)+2] to the reciprocal on both sides, 1 an=1 2+1 a(n-1).

1/an-1/a(n-1)=1/2

So {1 an} into a series of equal differences.

2) Find the general term of an.

Let cn=1 an, so cn is a series of equal differences, c1=1 a1=1, and the tolerance d=1 an-1 a(n-1)=1 2

CN=1+(n-1) 2

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

an=2/（n+1）

s2=a1+a2, so a1+a2=a3 Let the tolerance of the difference series be x, then a3=a2+x, i.e., a1+a2=a2+x, x=a1=1 2, a2=a1+x=1

1. A2+A3 = A1+D + (A1+2D) = 2*A1+3*D = 13 (1)S5 = 5*A1 + D + 2*D+3*D + 4*D) = 5*A1 + 10*D = 25

i.e. a1 + 2*d = 5 (2) is obtained by equations (1) and (2).

a1 = 11，d = -3。

2. S20 = 20*A1 + D*20*19 2= 220 - 3*190

Because a2 + a3 = 5

So a1 + d + a1 + 2d = 5, so 2a1 + 3d = 5

Because s5 = 20

So 5(a1 + a5) 2 = 20

So a1 + a5 = 8

So a1 + a1 + 4d = 8

So a1 + 2d = 4

So a1 = 2 and d = 3

So a10 = a1 + 9d = 2 - 27 = 25