On the question of the series, about the question of the series

Add an-1 on both sides of the recursive type

an+an-1=3 (an-1+an-2), an+an-1 is the n-1 term of the first proportional series with a2+a1=7 and the common ratio of 3, an+an-1=7*3 (n-2)...1)

Subtract 3an-1 from both sides of the recursive formula.

an-3an-1=-(an-1-3an-2), an-3an-1 is the n-1 term of the proportional series with the first term a2-3a1=-13 and the common ratio is -1, an-3an-1=(-13)*(1) (n-2)=13*(-1) (n-1).2)

1) by adding 3 times (2) and dividing by 4.

an=[7*3^(n-1)+13*(-1)^(n-1)]/4.

Question 1; If there are an bees on day n, then there are 5*an+an=6an on day n+1, that is, a(n+1)=6an. In this way, it becomes a proportional sequence, and on the sixth day there is 1 times 6 to the power of 6 to make 46,656 animals.

Question 2; Set 5 people to get A-2K, A-K, A, A+K, A+K, A+2K. If their sum is 100, then 5a = 100 and a = 20. According to the conditions in the question, 3(a+k) 3=a+k=2a-3k, k=5, and the minimum part is 20-2*5=10

1、a1=1,a2=a1x5+a1,a3=a2x5+a2……an=a(n-1)x5+a(n-1)=6xa(n-1).

So there is a series of proportional numbers, an a(n-1)=6.

So sn=a1x[(1-q n) (1-q)]=1x[(1-6 n) (-5)]=(6 n-1) 5, the sixth day arrives, i.e. n=7, so s7=(6 7-1) 5

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

By transforming.

2[a(n+1)]^2=a(n)

It can be obtained by using the properties of the recursive formula.

When n is equal to 1, 2, 3, 4, 5 respectively, it can be obtained.

2(a2)^2=a1

2(a3)^2=a2

2(a4)^2=a3

2(a5)^2=a4

2(a6)^2=a5

Available after finishing.

2^(1+2+4+8+16)*(a6)^32=a1=1a6)^32=2^-31

So a6=2 (-31 32).

2.Let the equation y=-x+b

Bringing (2,1) into the solution gives b=3

y=-x+3

Derivatives have not been learned.

f（1）=a+b=5/2

f（2）=2a+b=4

f（4）=4a+b=7

3.This function is symmetrical with respect to x=0.

In [0,2] single increase.

So f(0) minimum = -3

f(2) max=1

Because it's all positive.

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

n 2, there is an sn s(n-1) (an 1 an) 2 (a(n-1) 1 a(n-1)) 2

Simplification yields 1 an (an 1 an) 2 (a(n-1) 1 a(n-1)) 2 sn s(n-1).

Therefore (sn) s(n-1)) sn s(n-1))(sn s(n-1)) 1 an)an 1

and a1>0, and a1 s1 (a1 1 a1) 2, the solution is s1 a1 1, (s1) 1

So (sn) is the first series of equal differences with a term of 1 and a tolerance of 1.

At n 2, (sn) s1 (n-1) n, and sn>0, we get sn n

an sn s(n-1) n (n-1), and n 1 also holds the above equation.

Hence an n (n-1), n n

1 All (1) a(n+2) = (5 3) * a(n+1)-(2 3)*a(n)3a(n+2)=5a(n+1)-2*a(n)3a(n+2)-3a(n+1)=2a(n+1)-2a(n)3[a(n+2)-a(n+1)]=2[a(n+1)-a(n)][a(n+2)-a(n+1)] [a(n+1)-a(n)]=2 3, i.e., b(n+1) b(n) =2 3

b(n) is a proportional series.

b1=a2-a1=5/3-1=2/3

b(n) is the first proportional sequence with 2 3 and the common ratio 2 3 b(n) = (2 3) n

Finding a bn is simple.

bn=an+1-an;

an+2=5\3an+1-2\3an an+2-an-1=2/3*(an+1-an)

i.e. an+1-an=2 3*(an-an-1).=(2 3)(n-1)*(a2-a1) (Note: is n-1 power).

That is, bn=(3 2)n (to the nth power) can also be done with the eigenequation, and directly find an, and then find bn

Solution: bn 1 n(an 3) 1 [n(2n 2)] 1 [2n(n 1)] 1 2[(1 n) 1 (n 1)].

sn＝b1＋b2＋……bn

1/2[(1－1/2)＋(1/2－1/3)＋…1/n－1/(n＋1))]

1/2[1－1/(n＋1)]

n/[2(n＋1)]

Suppose there is an integer t satisfying the sum of SN t 36.

s(n＋1)－sn＝(n＋1)/[2(n＋2)]－n/[2(n＋1)]＝1/[2(n＋2)(n＋1)]＞0

Therefore, the series is monotonically increasing.

S1 1 4 is the smallest of the SN.

t/36＜1/4

i.e. t 9 and t n*

The maximum integer t is 8

It is known that the difference series an is 2n-1, let bn=1 n(an+3) sn=b1+b2+.bn whether there is a maximum integer t such that for any n there is sn greater than t 36 if there is t find t if there is no reason to state the reason.

(1) Bring b(n)=s(n)-s(n-1) into simplification to obtain 1 s(n)-1 s(n-1)=1 2

1 s(n)=(n+1) 2s(n)=2 (n+1) is obtained from the general term formula of the equal difference series

b(1)=1,bn=s(n)-s(n-1)=-2/(n(n+1)) n=2,3,4...

2) 81 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 3 is the third item in the thirteenth line.

b(13)=-1/91

So the common ratio is 2

It is obtained by the formula for summing the proportional series.

h(k)=b(k)+2b(k)+.2^(k-1))b(k)=b(k)*(2^k)-1)

-2/(k(k+1)))2^k)-1)

1）a1+3a2+3^2a3+……3 n-1an=n 3 subscripts are replaced with n-1.

a1+3a2+3^2a3+……3 n-2an-1=(n-1) is defeated by 3, and the two equations are subtracted to give 3 n-1an=1 3

So an=1 3 n

2)bn=n/an=n*3^n

sn=1*3+2*3^2+3*3^3+..n-1)3^(n-1)+n*3^n

At the same time, multiply 3 to get celery to the dry stove.

3sn= 1*3^2+2*3^3+..n-2)3^(n-1)+(n-1)*3^n+n*3^(n+1)

The difference between the two formulas gets:

2sn=(3+3^2+3^3+..3 n)-n*3 (n+1)3(1-3 n) (1-3)-n*3 (n+1) so. sn=(n/2-1/4)*3^(n+1)+3/4

Since 1+2+4+...512=1023<2009<1+2+4+..2 10 = 2047, so we know that a2009 is taken in a long string of 2s after the 11th 1.

That's easy to do: s2009 = 1 * 11 + 2 * 1998 = 11 + 3996 = 4007

67 minus 4 equals 63, indicating that the tolerance of the difference row is 3 or 9 or 7, and the slow first term of the new series is 4, and the unterm is 67,781 divided by (4+67=)71 and so on11, indicating that the new number series has 22 terms, then n=20