In the sequence an, a1 3, an an 1 2n 1 0

an+an-1+2n-1=0 an+n=-(an-1+(n-1)) an+n] [an-1+(n-1)]=-1 an+n is a proportional series, the first term a+1=4, q=-1 an+n=4*(-1) (n-1) an=4*(-1) (n-1)-n

When n is an odd number, sn=-(n+1)n 2-4

When n is even, sn=-(n+1)n 2

The first question should be to verify that a[n] is a series of equal differences.

When n=2, a[2]+a[1]+2*2-1=0 --n=3, a[3]+a[2]+2*3-1=0 --n=4, a[4]+a[3]+2*4-1=0 when n=2*k-1 (k is a natural number), a[2k-1]+a[2k]+2*(2k-1)-1=0 --2k].

When n=2*k (k is a natural number), a[2k]+a[2k-1]+2*(2k)-1=0 --2k+1].

Multiply all of the above formulas for even numbers by (-1) and add up all of the above equations.

a[2k]+a[1]+3+2*(k-1)=0a[2k]=-(2k+4)

When n=2*k+1 (k is a natural number), a[2k+1]+a[2k]+2*(2k+1)-1=0

a[2k+1]=-2k+3 ……

Second question: In the same way, the grouping can be added after classification. Omitted.

The original equation is deformed;

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

Therefore, (an+n) (an-1+n-1)=-1a1+1=4

Therefore, an+n=4*(-1) (n+1).

an=n+4*(-1) (n+1) From the general term of an, we can see that an is not a proportional series... Is there a problem with the question?

sn=(n+1)n 2+4 n is an odd number.

sn=(n+1)n 2 n is even.

You're typing the wrong question... a1=3, substituting the formula you gave to get a2=-6, a3=1....The first 3 items are not equal

Landlord, you're all wrong, I guess it's to verify that {an n} is a proportional series.

It is possible to argue that the slag is added with the state of the tired cavity.

a2-a1=2*1

a3-a2=2*2

a4-a3=2*3

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

Accumulate the upper and lower masking formulas to obtain: an-a1=2*(1+2+3+..n-1)an-a1=n(n-1)

That is: an=n(n-1)+1

a1=0a(n+1)=an+2n-1

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

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

a(n-1)-a(n-2)=2(n-2)-1a2-a1=2*1-1

Add up the above n-1 celery seeds to get:

an-a1=2*1+2*2+..2*(n-2)+2*(n-1)-(1+1+..)1）

2*[1+2+..n-1）]-n-1）

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

n-1)n(n-1)-(n-1)

n-1)^2

again a1=0

an=(n-1)^2

Solution: a(n+1)=an (2an+1).

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

1) Infiltrator an=1 an

1/a(n+1)

1 an=2 is a fixed value.

1 a1 = 1 3, the number series is the difference series with 1 3 as the first term and 2 as the tolerance.

1/an=1/3

2(n-1)=(6n

an=3/(6n-5)

When n=1, a1=3 (6-5)=1 also satisfies the general formula.

The general formula for a series of numbers is an=3 (6n-5).

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

a2-a1=2

a3-a2=2^2

a4-a3=2^3

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

The above formula is superimposed on the rubber brother.

an-a1=[2+2^2+2^3+..2 (n-1)]an-1=2[1-2 (n-1)] 1-2)=2 n-2, so Liang Lu's uproar an=2 n-1

Solution:1It is known that a(n+1)-an=2n, so:

a2-a1=2*1a3-a2=2*2a4-a3=2*3a5-a4=2*4。。。The addition of an-a(n-1)=2*9(n-1) will give a2, a3, a(n-1).These items are eliminated.

So we can get an-a1=2*1+2*2+2*3+2*(n-1)＝2*（1+2+3.。。n-1)=2*=n(n-1) and because a1=3, an=n(n-1)+3 is the application of the accumulation method!!

So an=n -n+3 and sn=(a1+a2+a3+..an=(1 +2 ++n)-1+2++n)+3n=(1 6)n(n+1)(2n+1)-n(n+1) 2+3n=(1 3)n(n +8) where there is a formula for the sum of n, that is, if an=n, then sn=n(n+1)(2n+1) 6, you can use it directly!! get an a(n-1)=(n+1) nSo, a2 a1=3 2a3 a2=4 3a4 a3=5 4...

a(n-2) a(n-3)=n-1 n-2a(n-1) a(n-2)=n n-1an a(n-1)=n+1 n, then a2, a3, a4...a(n-1) superposition an a1=(n+1) 2 knows a1=3 so an=3(n+1) 2 and when n=1, an=3(n+1) 2 3*(1+1) 2 3 so an=3(n+1) 2 This test of the cumulative multiplication!! This is the latter items to pay attention to, which are particularly prone to errors!!

If you don't know, you can ask again!!

From the condition a1 = 2, a2 = 5And there are: a2-a1=3*1, a3-a2=3*2, a4-a3=3*3,..

an-a(n-1)=3*(n-1), accumulatively, an-a1=3*(1+2+3+.n-1)=3n(n-1)/2.===>an=2+3n(n-1)/2.

It is verified that for any positive integer n, it is true, so the general term is an=2+3n(n-1) 2

The progressive formula can be adjusted as: a(n)+3n+7=2[a(n-1)+3(n-1)+7].

Let b(n)=a(n)+3n+7, then b(n)=2b(n-1), b(1)=a(1)+3*1+7=11

b(n)=b(1)*2 (n-1)=11*2 (n-1)then: a(n)=11*2 (n-1)-3n-7