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From the condition a2 = 6 3 = 2, a4 + a5 = -4 4 = -1 due to a4 + a5 = 2 * a2 + 5d (d is the difference number) to obtain d = -1

Hence an=4-n

Hence know bn=qn 2-1

So sn=q(1 2+2 2+.n^2)-n=q*n*(n+1)*(2n+1)/6-n

1> an=4-n;

As far as is known, a1+a2+a3=6, denoted as, a1+a2+.a8=-4, denoted as.

2}-=a4+a5+..a8=-10, denoted as.

Because an is an equal difference series, we can easily know that a1+a2+a3=3*a2=6, so a2=6 3=2. In the same way, from the result, a6 = -10 5 = -2.

Let the tolerance be x, then a6-a2=4x=-2-2=-4, so x=-1.

At the same time we know that a1=a2-x=3. Now that we know that a1=3 and the tolerance is -1, we can get an=3+(-1)*(n-1)=4-n according to the equation of the difference series.

2> qn refers to q to the nth power or q multiplied by n? I wrote the answers to both scenarios.

Case 1 (qn refers to q multiplied by n): according to the general formula of 1 >, bn=q*n 2-1 is obtained. Let cn=n 2 and dn be the sum of the first n terms of cn.

sn=q-1+q*2^2-1+..q*n^2-1=q+q*2^2+..q*n^2-n=q*dn-n。

Whereas. dn= 1 + 2^2 + n-1)^(2)+n^2=n*(n+1)*(2n+1)/6

So sn=q*n*(n+1)*(2n+1) 6-n.

Case 2 (qn to the nth power of q): According to the general formula of 1>, bn=(4-4+n)q n-1=nq n-1. Let cn=nq n, and the sum of the first n terms of cn is dn. where q n is to the nth power of q.

sn=q-1+2q^2-1+..nq^n-1=q+2q^2+..nq^n-n=dn-n

We just need to calculate the dn to know the sn.

dn= q + 2q^2 + n-1)q^(n-1)+nq^n

q*dn=0+ q^2 + n-2)q^(n-1)+(n-1)q^n+nq^(n+1)

q*dn-dn=(q-1)dn =-q-q^2-..q^(n-1)-q^n+nq^n=-q(q^n-1)/(q-1)+nq^(n+1)

Because the first n terms are proportional sequences with a common ratio of q.

So we get dn=-q(q n-1) [(q-1) 2]+nq (n+1) (q-1).

sn=dn-n=-q(q^n-1)/[(q-1)^2]+nq^(n+1)/(q-1)-n。

1 All an=3a(n-1)+2

an+1=3a(n-1)+3

an+1=3[a(n-1)+1]

Let bn=an+1

then bn=3b(n-1) (n 2)b1=a1+1=2

bn=2*3^(n-1)

an=bn-1

an=2*3 (n-1)-1 (n 2) and a1=1 satisfy the above equation.

So an=2*3 (n-1)-1

From the title, then: a(n+1)=3an+2, which is a type of equal difference series converted into an equal proportional series.

Add 1 to both sides of the equation at the same time, there is a(n+1)+1 = 3(an+1), let an+1 = bn

Then there is b(n+1) = 3 bn, so bn is a proportional series of 3, b1=a1+1=1+1=2, so it is derived from the general formula of the proportional series: bn=b1·3 (n-1)=2·3 (n-1).

And because bn = an + 1, the general term an = bn - 1 = 2·3 (n-1) -1

an=3a(n-1)+2

an+1=3a(n-1)+3=3[a(n-1)+1] so it is a proportional series with a common ratio of 3, a1+1=2

an+1=2×3^(n-1)

an=[2×3^(n-1)]-1

an=3a(n-1)+2, let an+k=3[a(n-1)+k], it can be obtained by comparing 2 formulas, k=1, so the condition can be written as an+1=3[a(n-1)+1], so (an+1) [a(n-1)+1]=3, a1=1, a2=3*1+2=5, so a1+1=2, a2+1=6, so an+1=2*3 (n-1), so an=2*3 (n-1)-1,

There's nothing wrong with the title.

How it doesn't feel right.

an+1=3(a+1)

It is a proportional series with 2 as the first term and 3 as the common ratio, an+1=2 times 3 to the n-1 power, so an=2 times (n-1 power of 3)-1

If you don't understand it, forget it.

Let a(n) = a(1) + n-1According to the title, a(b1)=a(1)+b(1)-1=4

So c(1)=4.If we also know that the tolerance is 1, then we can find the sum of the first ten terms of the series c(n) = 4 + 5 + 6 + ......13=85

The landlord may not understand the topic too much, so what a (b1) just bring b1 in.

an = sn-s(n-1)

n^2an - n-1)^2a(n-1)

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

n>1, both sides can be divided by (n-1).

Get. n-1)*a(n-1) =n+1)*an, i.e.: an a(n-1) =n-1) (n+1) so an = an a(n-1) *a(n-1) a(n-2) *a2 a1 * a1

n-1)/(1+n) *n-2)/n * n-3)/(n-1) *1/3 * 2

4 / n(n+1)

When n=1, a1 = 4 2 = 2 also agrees. Roll.

i.e.: an = 4 n(n+1)] is what is sought.

Hope it works, thanks

Solution: Set the tolerance to Brother D

a1+a2+a3=3a2=-3

a2=-1a1·a2·a3=8

a2-d)·a2·(a2+d)=8

a2 = -1 substitution, finishing, get d = 9

d = 3 or d = -3

When d=3, a1=a2-d=-1-3=-4

an=a1+(n-1)d=-4+3· (n-1)=3n-7d=-3, a1=a2-d=-1-(-3)=2an=a1+(n-1)d=2+(-3)(n-1)=-3n+5 The general term of the sequence is an=3n-7 or an=-3n+5 If an=-3n+5, then a1=2,a2=-1,a3=-4a1·a2=-2,a3=(4) =16,a1·a2≠a3 , contradicts the known, and therefore only.

a1=-4，an=3n-7

Ling an 0,3n-7 0,n 7 3

n is a positive integer, n 3, that is, the first 2 terms of the number series are negative, starting from the 3rd term of the burning Huai attack, and the subsequent terms are positive.

When n=1, s1=|a1|=|4|=4

n=2, s2=|a1|+|a2|=|4|+|1|=5n 3, sn=|a1|+|a2|+.an|-a1-a2+a3+..an

a1+a2+..an)-2(a1+a2)(-4+3n-7)n/2

3n²-11n+20)/2

Solution: (1) a2=a1+d=5, s4=a6+7=a1+5d+7=4a1+4 3 d 2, 3a1+d=7, and a1+d=5 solution gives a1=1, d=4 an=a1+(n-1)d=4n-3;

bn=2^[（an+3）/4]=2^n

2）anbn=（4n-3）×2^n

4（1×2+2×2^2+3×2^3+…+n×2^n）-3（2+2^2+2^3+…+2^n）

Let cn=n 2 n

sn=2+2×2^2+3×2^3+…+n×2^n①

2sn=2^2+2×2^3+…+n-1）2^n+n2^（n+1）②

Get sn=-2-2 2-2 3-...-2^n+n×2^（n+1）=n×2^（n+1）-（2+2^2+2^3+…+2 n)=n 2 (n+1)-(2-2 n 2) (1-2)=(n-1)2 (n+1)-2

tn=4（n-1）×2^（n+1）-8-3×1/（1-2）×（2-2^n×2）=（4n-7）×2^（n+1）-2

In the network, the symbol " " " is generally used to represent the power, e.g. 20 30 represents the 30th power of 20.

Let a1=k, then an=k*2 (n-1).

an is a positive integer.

k>=1 and is a positive integer.

So: a1a2a3....a30=k^30*2^(30*30/2)

2^(30*15)（∵k^30>=1)

That is, there are errors in the original question and there is no solution.

Change a condition: an is a drop-proportional sequence composed of positive numbers, and change from positive integers to positive numbers, then the solution is as follows:

a1a2a3...a30=(a1a30)^15=20^30=(20²）^15

a1a30=20²

a3a30=a1*2²*a30=4*a1a30=4*20²=40²

then a3a6a9....a30=(a3a30)^5=(40²）^5=40^10

a1a2a3...a30 = (a1a30) drops to the 15th power = 20 drops to the 30th power So a1a30 = 20 drops squared.

a3a6a9...a30 = (a3a30) drops to the 5th power = (a1 * q 2a30) drops to the 5th power = (a1a30) drops to the 5th power * q drops to the 10th power = 20 drops to the 10th power * 2 drops to the 10th power = 40 drops to the 10th power.

an+1/an=n/（n+2）

an/an-1=(n-1)/(n+1)

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

a3/a2=2/4

a2/a1=1/3

Multiply all of the above to get.

an/an-1)*(an-1/an-2)*.a3/a2)*(a2/a1)*a1 =(n-1)/(n+1)*(n-2)/n*..2 4*1 3*2k, you can push an=4 (n+1)(n+2).

Just stack it and you're good to go.

an/an-1=n-1/n+1

an-1/an-2=n-2/n

a2/a1=1/3

Multiply left and right.

an/a1=(n-1)！/[(n+1)!2] So an=4 n(n+1) n>1 a1=2

Forehead... The one on the first floor must have misunderstood the meaning of the landlord.

The an+1 on the left should be a(n+1).

Divide directly by Ah and then 1+1 an=n (n+2).

Then the right side of the equation is the common ratio q.

A1 Got it, and then the general term is easy to ask for.

n*(n+1) 4/4, from the problem an an+1=(n-1) (n+1), and so on to a2 a1=1 3, and then multiply, remove the same numerator and denominator, to get an a1=2 n*(n+1), because a1=2, so, an=4 n*(n+1).