It is known that the sequence of an is a1 1, an a n 2 1 n is an even number or 1 a n 1 n is an odd n

a1=1 a2=a1+1=2 a3=1/a2=1/2

None of the above derivations make an<0

an>0

n is an even number an=a(n 2)+1>1

n is an odd number an=1 a(n-1) a(n-1) is an even term a(n-1)>1 an<1

an=30/19>1

n is an even number. a(n/2)=(an)-1=30/19-1=11/19<1

a(n2) is an odd number.

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

a((n-2)/2)=1/a(n/2)=19/11>1

a((n-2) 2) is an even number.

a((n-2) 4)=a((n-2) 2)-1=8 11<1 odd terms.

a((n-2) 4-1)=a((n-6) 4)=1 a((n-2) 4)=11 8>1 Even terms.

a((n-6) 8)=a((n-6) 4)-1=3 8<1 odd terms.

a((n-6) 8-1)=a((n-14) 8)=1 a((n-6) 8)=8 3>1 Even terms.

a((n-14) 16)=a((n-14) 8)-1=5 3>1 even terms.

a((n-14) 32)=a((n-14) 16)-1=2 3<1 odd-numbered terms.

a((n-14) 32-1)=a((n-46) 32=1 a((n-14) 32)=3 2>1 Even terms.

a((n-46) 64)=a((n-46) 32)=1 2<1 odd-numbered terms.

a((n-46) 64-1)=a((n-110) 64)=1 a((n-46) 64)=2>1 even-numbered terms.

a((n-110)/128=1/a((n-110)/64)-1=1=a1

n-110/128=1

n=238 If you don't see it clearly, check it again.

a238=30/19>1

a119=a238-1=11/19<1

a118=1/a119=19/11>1

a59=a118-1=8/11<1

a58=1/a59=11/8>1

a29=a58-1=3/8<1

a28=1/a29=8/3>1

a14=a28-1=5/3>1

a7=a14-1=2/3<1

a6=1/a7=3/2>1

a3=a6-1=1/2<1

a2=1/a3=2>1

a1=a2-1=1

1a1=1,a2=1/2a1+1=3/2

2.When n is an even number of 2, an=1 2a(n-1)+(n-1)=1 2[ a(n-2)-2(n-2)]+n-1=1 2a(n-2)+1

So let n=2k(k z), a2k=1 2a2(k-1)+1

So a2k-2=1 2a2(k-1)-1=1 2[a2(k-1)-2].

It is a proportional series with the first term -1 2 and the common ratio of 1 2.

a2k-2=-1/2*(1/2)^(k-1)=-1/2)^k

So the general formula for even terms in is an=-(1 2) (n 2)+2 (n is even).

is an odd number, an=(a(n+1)-n)*2=*2=-(1 2) [n-1) 2]-2(n-2).

a1+a3+..a99=-1-(1/2)^1-(1/2)^2-..1/2)^49-2-6-10-..198+4*50=

(1) According to known conditions:

b(n+1)=a2(n+1)-2=a[(2n+1)+1]-2=1 2a(2n+1)+2n+1-2=1 2a(2n+1)+2n-1=1 2(a2n-2x2n)+2n-1=1 2a2n-2n+2n-1=1 2a2n-1=1 2(a2n-2) (In this process, 2n+1,2n need to be regarded as a whole, which are odd and even).

So b(n+1) bn=1 2=equivalence, b1=a2-2=-1 2, bn=(1 2) n-1

2) According to equation (1), bn=a2n-2=(1 2) n-1, so a2n=(1 2) n+1

s=a2+a4+a6+……a100=1/2+1+(1/2)^2+1+(1/2)^3+1+……1/2)^50+1=51-(1/2)^50

Hopefully there are no miscalculations.

bn=a(2n+1)+4n-2

b(n+1)=a(2n+3)+4n+2

a(2n+2)-2(2n+2)+4n+2=1/2a(2n+1)+2n-1

1/2[a(2n+1)+4n-2]

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

The number series is an equal proportional series, and the common ratio is 1 2

b1=a3+2=a2-4+2=1/2a1+1-2=-1/2bn=-(1/2)^n

bn=a(2n+1)+4n-2

a(2n+1)=bn-4n+2=-1/2^n-4n+2s=a1+a3+a5+..a99

Solution] 4, 5, 32 three.

a6=1 ==> a5=2 ==> a4=4 ==> Two cases:

a3=1 ==> a2=2 ==> a1=4, i.e. m=4;

a3=8 ==> a2=16 ==> Two cases:

a1=5, i.e. m=5;

a1 = 32, i.e. m = 32.

To sum up, there are three possible values of m: 4, 5, and 32.

a2=1/2*a1+1=

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

Let cn=bn-2=a(2n)-2

c1=a2-2=

c(n-1)=a(2n-2)-2, i.e., a(2n-2)=c(n-1)+2cn=a(2n)-2=1 2*a(2n-2)+1-2=1 2*[c(n-1)+2]+1-2=1 2*c(n-1).

Therefore, cn is the first proportional series with c1= and the common ratio q=1 2.

The general formula is:

cn=c1*q^(n-1)=

The sum of the first n terms is.

sn=c1*(1-q) (1-q)=1 (2 n)-1, so bn=cn+2=2-1 (2 n).

a(2n+1)=a(2n)-2(2n)=a(2n)-4na(2n)+a(2n+1)=2a(2n)-4n=2(cn+2)-4n=2cn-4(n-1)

So. tn=a1+a2+a3+……a（2n+1）=a1+(a2+a3)+.a(2n)+a(2n+1)]=1+2c1+..2cn-4(n-1)]

1+2(c1+c2+..cn)-4[1+2+..n-1)]=1+2sn-4n(n-1)/2

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

Please check it out. Be.

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

Or (1 2an)+n

This question is wrong, don't do it:

an+1=1 2(an+n), which is an=n-2 in the condition, a1=1, brought in, a1=1-2=-1....Contradictory.