A series of questions in the first year of high school. Big brother and sister help. Hurry

Abdication subtraction, the teacher has taught it, that is, n is to become (n-1) in the subtraction with the original formula, sn and s(n-1) are equal to an, you know the next.

sn+1(subscript)=3(n+1)squared-3(n+1)+1,an+1=sn+1-sn=4n-1=4(n+1)-5, so an=4n-5(n starts at 2!) ), a1=0 is written separately.

When n>1, sn=2n2-3n+1

sn-1=2(n-1) 2-3(n-1)+1 is subtracted by the above two equations to obtain:

an=4n-5, when n=1, it does not conform to the formula, and because a1=s1=0, so: an=0, n=1

4n-5 ，n>1

This is segmented).

Square that 2n -3n+1

Swap for 1 (2n-1)*(n-1) by multiplying the crosses

It is then reduced to 1 n* (1 2 n-1 minus 1 n-1).

sn=2n squared - 3n+1

sn-1=2(n-1) 2-3(n-1)+1 subtracted to an=4n-5 n>1

But when an=0 n=0

s5=a1+a2+a3+a4+a5=5a3 15 so a3 3

s4 10 then -s4 -10

s4+a5=s5

a5 = s5 - s4 = s5 + (-s4) 15 + (-10) = 5 i.e. a5 5

a3+a5=2a4≤3+5

So a4 4

That is, the maximum value of a4 is 4

Haha, I'm also a freshman in high school, I'll give it a try.

bn=2 (an) bn+1 = 2 [a(n+1)], then bn+1 bn (quotient between the last term and the former term of the number bn) =2 [a(n+1)-an]=2 (3n-2-3n+5)=2 3=8

For the first proportional sequence b1=2 a1, then the sum is also the sum of the proportional sequence, and the summing formula is OK.

And then the second way:

Is that most positive period the smallest positive period?

If so, use the cosx double angle formula y=cos(2x) then the minimum positive period is t=2 2= .

I don't know if it's right or not, I hope the landlord uses pulling

The sum of the first n terms of bn is n + n

Solution: an=4n-1

a1+a2+..ak

4×1-1)+(4×2-1)+.4×k-1)=4(1+2+..k)-1×k

2k²+kbk=(2k²+k)/k=2k+1

b1+b2+..bn

2×1+1)+(2×2+1)+.2×n+1)=2(1+2+..n)+1×n

n + n so the sum of the first n terms of bn is n + n

(1) Proof : Because an=bn(1 b1+1 b2+..1/bn-1) （n≥1）

So a(n+1)=(bn+1)(1 b1+1 b2+.1/bn) （n≥1）

So (an+1) a(n+1)=[bn(1 b1+1 b2+..1/bn-1)+ 1]/（bn+1）(1/b1+1/b2+..1/bn)

bn(1/b1+1/b2+..1/bn-1+ 1/bn-1/bn)+1]/（bn+1）(1/b1+1/b2+..1/bn)

bn(1/b1+1/b2+..1/bn-1+ 1/bn)-1+1]/（bn+1）(1/b1+1/b2+..1/bn)

bn(1/b1+1/b2+..1/bn-1+ 1/bn)/（bn+1）(1/b1+1/b2+..1/bn)

bn/b(n+1)

2) Proof : an+1 a(n+1)=bn b(n+1) from (1) because bn=2 (n-1).

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

2^(n-1）/2x2^(n-1)

So (1+1 a1)(1+1 a2)...1+1/an)=[(1+a1）/a1][(1+a2)/a2]..1+an)/an]

(1+a1)(1+a2)..1+an)]/(a1a2...an)

1+an)/[a1x2^(n-1)]

And because bn=2 (n-1), a1=1, so b1=1, so the original formula = 1+2 (n-1)(1 b1+1 b2+.1/bn-1)/2^(n-1)

1+(1/b1+1/b2+..1/bn-1) (n∈n*)

1+[1/2^0+1/2^1+..1/2^（n-2）]

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

3-(1/2)^(n-1) (n∈n*)

Because (1 2) (n-1)>0 (n n*).

So 3-(1 2) (n-1)<3<10 3 i.e. (1+1 a1)(1+1 a2)...1+1/an)<10/3

(1) When n=1, a1=1, b1=1, b2=2, a2=2 does not satisfy the equation (is this what you want to represent bn=2 (n-1) means).

When n>=2, an bn=1 b1+1 b2+.1 bn-1 is substituted into an+1 bn+1=1 b1+1 b2+.1 bn-1+1 bn=an bn+1 bn, variant (an+1) a(n+1)=bn b(n+1).

Freshman in high school?? How come I haven't studied?

1 Is the denominator of the formula 3(a(n-1)+2a(n-2)) 3(a(n-1)+2a(n-2)) or 3?

Solution: The relationship between the two companies and the quarterly number of grinding cover x:

A: y1 = B: y2 = 6 + 2 (x 4-1) =

When x>6

When y1 > y2, >

x<6 Answer: Company B is high in the first six Ming years, and A has a high salary after six years.