The number series An where a2 6 is known, and an 1 an 1 an 1 an 1 n , find the general formula for

Simplification is obtained first.

n-1)an+1=(n+1)an-(n+1) Here it is easy to introduce a1=1 and then convert it.

n-1)*(an+1- (n+1))=(n+1)*(an-n)Let bn=an-n

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

b2=4b3=b2*3/1

b4=b3*4/2

bn=bn-1*(n)/(n-2)=...=4*(3*4*5*..n)/(1*2*3*..n-2))=4*n*(n+1)/2=2n^2-2n

So. an=2n^2-n（n>=2）

Verify that n=1 satisfies the general formula when a1=1 satisfies the general term.

So. an=2n^2-n

Because an+1 + an - 1) (an+1 - an + 1) = n, so.

an+1)+(an-1)=n(an+1)-n(an-1) i.e.

n-1)(an+1)=(n+1)(an-1), so (an+1) (an-1)=(n+1) (n-1).

So (an) (an-1) = n (n-1).

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

an-2)/(an-3)=(n-2)/(n-3)

a3/a2=3/2

a2/a1=2/1

an=(an)/(an-1)*(an-1)/(an-2)*(an-2)/(an-3)*.a3/a2)*(a2/a1)*a1

n/(n-1)]*n-1)/(n-2)]*n-2)/(n-3)]*3/2)*(2/1)*a1

n*a1 and a2=6, and a2 a1=2 1, so a1=3

Finally, an=3n

Refer to it. 1）an+1-an=2^n

an-an-1=2 (n-1), a2-a1=2 1

an+1=an+1-an+an-an-1+, a3-a2+a2-a1+a1

2^n+2^（n
^1+1

2^（n+1）-1

an=2 n-1

2) From (Nendong 1): an=2 n-1

bn = n(2 n-1) = n2 n-n

sn=n2 n-n+(n-1)2 (n-1)-(including the first n *2 2-2+1*2 1-1

n2^n+（n-1）2^（n
*2^2+1*2^1-（n+（n
+1）

n2^n+（n-1）2^（n
*2^2+1*2^1-n（n+1）/2

Let b=n2 n+(n-1)2 (n * 2 2 + 1*2 1

Then nb=n2 (n+1)+(n-1)2 n *2 3+1*2 2 (dislocation subtraction).

nb-b=n2^（n+1）-2^n-2^（n
^3-2^2-2^1

n2^（n+1）-（2^（n+1）-2）

n-1）2^（n+1）+2

Sosn=(n-1)2 (n+1)-n(n+1)Aberdo 2+2 praise,

an+1=an+6(an-1)=> sock tremor (an+1)-3an=-2an+6(an-1)=-2[an-3(an-1)]=order haoqin bn=an-3(an-1)then: potato defeat bn=-2(bn-1) b1=-10=>bn=(-2) (n-1)*(10)=-10*(-2) (n-1)=>an-3(an-1)=-10*(-2) (n-1)3(an-1)- 9(an-2)=[10*(-2) (n-1)]*3...3^..

Split 2an + 3 5 n into 2 (an - 5 n) +2 5 n + 3 5 n = 2 (an-5 n) +5 (n+1) and then move 5 (n+1) to the left, so an+1 - 5 (n+1) =2 (an - 5 n) =2 2 (an-1 - 5 (n-1)) 2 n (a1 - 5 1) =2 n Look at the left and right of the most Tongdan, and get an+1 = 2....

The numerator and denominator can be reversed.

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

So the sequence 1 an is a series of equal differences with 2 as the first term and 2 as the tolerance.

1/an = 2n

So an = 1 2n

If there's anything you don't understand, you can ask me.

As can be seen from the title, all the items in the sequence are positive. Taking the reciprocal on both sides of the equation an+1=2an an+2 at the same time can be obtained as an equal difference series, so that the general term formula of the number series can be found first, and then the general term formula can be obtained.

Provide your ideas, and please complete the specific process yourself.

The general formula an=-1 n, please refer to the following process.

by an=an-1

2n gets an-an-1=2n

Then when n faction Cong Kuan Zheng Fan 2.

There is a2-a1=2x2

a3-a2=2x4

an-an-1=2n

Add the above n-1 equations.

an-a1=2(2+3+4+....)+n)

an-a1=2[(n-1)(n+2) dust bright2]an-a1=n2+n-2

So an=n2+n-1

When n 1, a1=1+1-1=1 also fits the above equation.

So the general formula.

is an=n2+n-1

Because a1=1,an=3 (n-1)·a(n-1)(n2,n n*) is obviously an 0

Therefore, lnan=ln[3 (n-1)·a(n-1)]=ln3 (n-1)+lna(n-1)=lna(n-1)+(n-1)ln3

So lnan-lna(n-1)=(n-1)ln3, so lna2-lna1=ln2

lna3-lna2=2ln3

lna4-lna3=3ln3

lnan-lna(n-1)=(n-1)ln3 is superimposed to obtain lnan-lna1=[1+2+..n-1)]ln3=[n(n-1)/2]ln3

So lnan=[n(n-1) 2]ln3+lna1=[n(n-1) 2]ln3+ln1=[n(n-1) 2]ln3=ln[3 [n(n-1) 2]].

Therefore an=3 [n(n-1) 2].