Satisfaction bonus Find the first 4 terms of a n series a 1 1, a n 1 a n 1 a n

If the question is: Find the first 4 terms of the series a(1)=1, a(n)+1=a(n)+1 a(n), that's right!

Then a(n)=1,a(1)=a(2)=a(3)=a(4)=1If the problem is: find the first 4 terms of the series a(1)=1,a(n+1)=a(n)+1 a(n), it feels like this!

a(1)=1

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

a(3)=2+1/2=5/2

a(4)=5/2+1/(2/5)=5/2+2/5=29/10

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

Subtract a(n) from both sides

1/a(n)=1

a(n)=1

Conclusion: All 1

Trouble....Subtract (1, 5) 2 from each side of the equation

It's minus twice....See clearly.

an-(1+√5)/2=[(1-√5)a(n-1)+2]/2a(n-1)

an-(1-√5)/2=[(1+√5)a(n-1)+2]/2a(n-1)

Divide the two formulas to get.

an-(1+√5)/2]/[an-(1-√5)/2]=[1-√5)a(n-1)+2]/[1+√5)a(n-1)+2]=[1-√5)/(1+√5)][a(n-1)-(1+√5)/2]/[a(n-1)-(1-√5)/2]

If you don't understand it, then I don't have to do it....)

Apparently proportional.

The common ratio is (1-5) (1+ 5), i.e. (-3+ 5) 2

So [an-(1+ 5) 2] [an-(1+ 5) 2]=[3+ 5) 2] n

Simplify it again, and get it.

an=[(1+√5)+(1-√5)[(5-3)/2]^(n-1)]/2[-1+[(5-3)/2]^n]

The landlord can be checked by the calculation calculator....Don't press the wrong number.

From the meaning of the title, an+1

an2, so count Hu Kailie {an

It is a series of equal differences with 2 as the common and closed to make a crack, and a3

8, so a6

a33d=8+6=14, so the answer is: 14

It is known that a (n+1) = a n+4

then a (n+1)-a n=4

So it's a series of equal differences with a tolerance of 4.

First term = a1 = 9

Therefore, an = 9 + 4 (n - 1) = 4n + 5

So an= (4n+5).

If an0n-5)(n+1)>0

Solve n<-1 or n>5

Because n n so n > 5

Hope it helps you o(o

a²（n+1)=a²n+4

The definition of equal difference is known to be a series of equal differences.

First term = 9, tolerance = 4

Therefore, a n = 4n + 5

Therefore an = root number under (4n+5)5

a²（n+1)=a²n+4

a²（n)=a²(n-1)+4

a²（n-1)=a²(n-2)+4

a²（n-2)=a²(n-3)+4

a²（3)=a²（2)+4

a²（2)=a²（1）+4

The left and left are added together, and the right and right are added together.

A(n+1)=a(1)+4*n

a1 = 3 substitution.

a²（n+1)=9+4*n

an = root number (4n+5).

Root number (4n+5), 5 or n<-1 (rounded).

Take the reciprocal. 1 a(n+1)=(an+1) an1 a(n+1)=1+1 an

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

1 Jing Qin pretends to be an isoluminous stove difference series, d=1

a1=11 an=1 a1+(n-1)d=n, so an=1 n

Both sides take the reciprocal to get the state closed 1 a(n+1)=1 an+1, which is a series of equal difference numbers to judge the closed school.

The excavations are an=1 n

1), a1|>2, So, A does not belong to M

2), 00 can actually be guessed;

So, a(n+1)>an, which is a bounded sequence, is now used to disprove it, assuming |an|<2 is true, and an is a monotonically bounded number series, and there must be a limit, and when n tends to positive infinity, the limit of an is b;

There must be, b = b 2+a1, i.e. b 2-b+a1=0 has a real root, and delta = 1-4a1<0, so b 2-b+a1 has no real root, so contradict each other, hence |an|<2 is untenable, i.e., A does not belong to M.

2. The general formula for finding the sequence an is 2 b(n+1)-bn=2bn+2-bn=bn+2 b(n+1)+ plus an-a2=2 3+2 4+2^n-2*(n-2) an-a2=2^3[1-2^(n-2

1an is a proportional series, since a1=1 4 and the common ratio q=1 4, we get an=(1 4) n

2 from the meaning of the title can be obtained b(n+1)+2=3log1 4a(n+1)b(n)+2=3log1 4a(n).

Subtracting the two formulas yields b(n+1)-b(n)=3(log1 4a(n+1)-log1 4a(n))=3log1 4【a(n+1) an】=3log1 4(1 4)=3

And b1+2=3log1 4(a1)=3 so b1=1 so bn is a tolerance of 3, and the equal difference series 3cn of the prime minister is 1 can be seen as the sum of an equal difference series and an equal ratio series.

Therefore, the sum of his first n phases can be found separately and then added.

Because the number column is proportional to the series, a1 1 4, q 1 4

So, an=1 4 (1 40 (n-1)=(1 4) n

Because, bn+2=3log1 4an=3log1 4(1 4) n=3n

So bn=3n-2

Because, b(n+1)-bn=3

Therefore, bn is a series of equal differences.

Because, the sequence satisfies cn=an·bn

So, cn=(3n-2) (1 4) n

sn=1/4+4×(1/4)^2+7×(1/4)^3……+3n-2)×(1/4)^n

sn/4= (1/4)^2+4×(1/4)^3……+3n-5)×(1/4)^n+(3n-2)(1/4)^(n+1)

Subtract the upper and lower formulas (dislocation subtraction):

3sn/4=1/4+3×(1/4)^2+3×(1/4)^3+……3 (1 4) n (3n-2)(1 4) (n+1) (sum in the middle with a proportional sequence).

1/4 +1-(1/4)^n－(3n-2)(1/4)(n+1)

3sn=5-(3n+2)×(1/4)^n

So, sn=[5-(3n+2) (1 4) n] 3

a(n+1)+x(n+1)+y=2a(n)+3n-4+x(n+1)+y=2[a(n)+xn+y],3n-4=xn+y-x,x=3,y=-1.

a(n+1)+3(n+1)-1=2[a(n)+3n-1],a(n)+3n-1} is the first proportional sequence with a(1)+3-1=1 and a common ratio of 2.

a(n)+3n-1=2^(n-1),n=1,2,..

a(n)=2^(n-1)+1-3n,n=1,2,..

a(n+1)-a(n)+3=2 (n)+1-3(n+1)-2 (n-1)-1+3n+3=2 (n-1) is a proportional sequence with the first term 1 and the common ratio of 2.

s(n)=1+2+..2^(n-1)+n-3[1+2+..n]=2 (n)-1+n-3n(n+1) 2 ,..

n=1,2,3,4,a(n)《0.

When n>=5, a(n)>0

t(1)=-s(1)=-2-1+1-3]=1,t(2)=-s(2)=-2 2-1+2-9]=4,t(3)=-s(3)=-2 3-1+3-18]=7,t(4)=-s(4)=-2 4-1+4-30]=11,n>=5,t(n)=s(n)-2s(4)=2 (n)-1+n-3n(n+1) tangerine2-2[2 4-1+4-30]=2 (n)+n-12-3n(n+1) 2