It is known that in the sequence an 1 1, an a n 1 2 n

1) Using the counter-proof method, assuming that it is a proportional series, then a(n+1) an=c

ana(n+1)=2 n

So we get a(n+1) 2=2 nc, an 2=2 n c

a(n+1) 2=2 (n+1) c

So 2 nc=2 (n+1) c

c= 2, so an=(2) (n-1), a(n+1)=(2) n

ana(n+1)=2 n2, contradictory.

From a(n+1)a(n+2)=2 (n+1), a(n+2) an=2 is obtained

and a1a2=2, a2=2

So when n is odd, an=an a(n-2)*a(n-2) a(n-4)*a3/a1*a1=2^[(n-1)/2]

When n is even, an=an a(n-2)*a(n-2) a(n-4)*a4/a2*a2=2^(n/2)

2) When n is an even number, the method of group summation is adopted.

The odd term is a proportional sequence of the first term 1 common ratio to the number n 2, and its sum is 1*[1-2 (n 2)] (1-2)=2 (n 2)-1

The even term is a proportional sequence of the first term 2 common to the number n 2, and its sum is 2*[1-2 (n 2)] (1-2)=2(2 (n 2)-1).

Therefore sn=3(2 (n 2)-1).

When n is an odd number, the odd term is an equal proportional sequence of the first term 1 common to the number of 2 terms (n+1) 2, and its sum is 1*[1-2 ((n+1) 2)] (1-2)=2 ((n+1) 2)-1

The even term is a proportional sequence of the first term 2 common terms (n-1) 2, and its sum is 2*[1-2 ((n-1) 2)] (1-2)=2 ((n+1) 2)-2

Therefore sn=2 ((n+3) 2)-3

Or in this way, when n is odd, n-1 is even, sn=s(n-1)+an=3[2 ((n-1) 2)-1]+2 ((n-1) 2)=2 ((n+3) 2)-3

Solution: Because an*a(n+1)=2 n

Replace n with 2n.

a(2n)*a(2n+1)=2^(2n)

i.e. a(2n-1)*a(2n)=2 (2n-1), so a(2n+1) a(2n-1)=2

Therefore, the series is a proportional series with a1 as the first term and 2 as the common ratio.

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

In the same way, replacing n with 2n+1 gives a(2n+1)*a(2n+2)=2 (2n+1).

That is, a(2n)*a(2n+1)=2 (2n) is divided by the two formulas to obtain a(2n+2) a(2n)=2

It can be found that the series is a proportional series with a2=2 as the first term and 2 as the common ratio.

a(2n)=2^n

anan+1=(1/2)^n

anan-1=(1/2)^n-1

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

a2*a1=1/2^1

an*a1==(1/2)^n-1*(1/2)^n-2*(1/2)^n-3……*1/2^1=1/2^[(1+n-1)(n-1)/2]=1/2^[n(n-1)/2]

an=1/2^[n(n-1)/2]

a2n=1/2^[2n(2n-1)/2]

a(2n-1)=1 2 [(2n-1)(2n-2) 2] series and are proportional series.

t2n=[1-1/2^[2n(2n-1)/2*1/2]]/[1-1/2]=ok

64*t2n*a2n 3*(1-k*a2n) can be brought into the solution.

I've seen a similar topic, FYI:

In the sequence, a1=1 2, a(n+1)=an 2+an, and the verification is: 1 (a1+1)+1 (a2+1)+1/(an+1)<2

Proof] a(n+1)=an(an+1), take the reciprocal, 1 a(n+1)=1 [an(an+1)], and the right split term: 1 a(n+1)=1 an-1 (an +1).

1/(an +1)= 1/an-1/ a(n+1)s=1/(a1+1)+1/(a2+1)+.1/(an+1)(1/a1-1/a2)+ 1/a2-1/a3)+ 1/a3-1/a3)+…1/an-1/ a(n+1))

1/a1-1/ a(n+1)

again a1 = 1 2, an increases, so s=2-1 a(n+1)<2

a(n+1)a(n)=a(n+1)-a(n) both sides at the same time divided by a(n+1)a(n), get:

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

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

So it is a series of equal differences with a tolerance of -1.

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

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

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

Add the two sides of the above n equations to obtain:

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

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

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

So. a(n)=-1/n

2*a(n+1) 2 n=an 2 (n-1)+1 order an 2 (n-1)=bn b1=1

2b(n+1)=bn+1 2b(n+1)-2=bn-1 makes bn-1=cn

c1=0 cn is the first proportional series of 0, and it is also explained that cn is a constant series of 0, that is, bn-1=0 an 2 (n-1)-1=0 an=2 (n-1).

①.a1=1,a(n+1)=2an (an+2), take the reciprocal to get: 1 a(n+1)= (an+2) (2an)

That is, 1 a(n+1)=1 an+1 2, so it is a series of equal differences with the first term 1 and the tolerance is 1 2, 1 an=1+(n-1) 1 2, an=2 (n+1)

bn=an •a(n+1)=4/[(n+1)(n+2)]=4/(n+1)-4/(n+2),tn=b1+b2+……bn=4[(1/2-1/3)+(1/3-1/4)+ 1/(n+1)-1/(n+2))]

4[1/2-1/(n+2)]=2-4/(n+2)< 2.

a(n+1)=an (1+2an) (take the reciprocal of both sides)1 a(n+1)=(1+2an) an

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

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

Therefore, it is a series of equal differences with 1 a1 = 1 as the prime minister and d = 2 as the tolerance: 1 an=1+2(n-1)=2n-1

So an=1 (2n-1).

Because an+2an*a(n-1)-a(n-1)=0, an-a(n-1)=-2an*a(n-1)an-a(n-1)} {an*a(n-1)}=-21 an)-=2

Because bn=1 an

So bn is a series of equal differences.

And the tolerance is 2bn=1 a1+2(n-1)=2n-1=1 an, so an=1 (2n-1).

If you don't understand, you can ask!! Thank you!