A simple problem in mathematics about sequences?

There is no a0 term in the sequence, your maximum abscissa is in between, there are only 10 numbers on the left, you symmetrical the ten numbers on the left to the right, and only when the maximum value is is you can get 20. Because there is no term before a1, you can only symmetrically pass a1, and the maximum coordinate you use is multiplied by 2, which is to project the number before a1. Your method is like if I draw a straight line with a downward slope (a series of equal differences) and then take the area equal around the zero crossing, but you actually ignore that the left side of the 1 is no longer counted.

I'm talking a lot of nonsense, because the series are discrete points, so sometimes you have to make some changes with a continuous function solution, and then ask if you have a question.

If b(n) = na(n), then b(n+1) = (n+1)a(n+1).

n+1)a(n+1)=na(n）

So there is b(n+1)=b(n).

That is, if you replace it with a variable, all the elements in the sequence b(n) are the same.

The bn=n*an you wrote yourself, this is the general term of the number series bn, so obviously b(n+1)=(n+1)a(n+1)ah.

At the beginning of the year after A1.

After 2 years a-

After 3 years a- x-

10 years later a- x -

1) If in 10 years' time, the per capita housing area in the region doubles compared to today, how much more old housing should be demolished every year?

a-〖 x⋯-〖=2a

calculated as x=2) According to the demolition speed of (1), how many years will it take to demolish all the old rolling houses that need to be demolished? Year.

Therefore, it will take more than 18 years to dismantle it.

Because s9=s17, a10+a11+a12+.a17=0, a10+a17=a11+a16=a12+a15=a13+a14=a1+a26=0 (lower corner coordinates and formula).

and a1 = 25 a26 = -25, from which it can be obtained that the tolerance of the equal difference is -2, so that the first n terms and the maximum should be positive, a13 = 1, a14 = -1 so the first 13 terms and the maximum, the maximum value is 169

1. That symbol represents a series of numbers, and the source of blue lead is an early whole, and it is not said that this series must be an infinite series of numbers up to n, so it is not right.

2. Not necessarily regretful, some sequences are poor. Only up to item n.

3. an+1 - an=1 (n+1) is always positive, so it is monotonically increasing.

an=sn - sn-1 = 2^n -1 - 2^(n-1) -1] =2^n - 2^(n-1)

2 n x (1-1 lingsheng2) =2 (n-1).

I think it's right for Yuanzhao, orange rent.

The idea is to simplify first and then use the superposition method or the stacking method. n*an+1=n*an+2an+n an=(n-1) 2 and because a1=1

a1=1an= (

n-1)/2 n>1

2a(1)=2s(1)=[a(1)] 2 + 1 - 4, 0 = [a(1)] 2 - 2a(1) -3 = [a(1)+1][a(1)-3], a(1)=-1(round), a(1)=3

2s(n+1)=[a(n+1)] 2 + n+1) -4,2a(n+1)=2s(n+1)-2s(n)=[a(n+1)] 2 - a(n)] 2 + 1,0=[a(n+1)] 2 - 2a(n+1) +1 - a(n)] 2 = [a(n+1)+a(n)-1][a(n+1)-a(n)-1], a(n+1)=-a(n)+1 or a(n+1)=a(n)+1, if a(n)>=1, then only a(n+1)=a(n)+1, is the first a(1)=3 and the tolerance is 1. a(n)=3+(n-1)=n+2.

If 0=1, a(n)=n+2

Solution: 2sn=an 2+n-4

Let n=1 have 2s1=2a1=a1 -3

Get: a1=3

2s(n-1)=a(n-1)^2+n-5

Subtract the two formulas, and there is.

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

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

Since a1=3, and each item is positive, then the tolerance d>0, and the square of the arithmetic on both sides of an 3 is, yes.

an-1=a(n-1)

Deliberately a series of equal differences.

an=3+(n-1)=n+2

∵2sn=an^2+n-4

2s1=a1^2-3

a1 = 3 or a1 = -1 (rounded).

2sn-1=an-1^2+n-5

Subtract the two formulas to obtain: 2(sn-sn-1)=an 2-an-1 2+1, i.e., 2an=an 2-an-1 2+1

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

an-1=a(n-1)

an-a(n-1)=1

An is a series of equal differences, and an=3+(n-1)*1=n+2 tells you that in fact, the general formula for finding a series of numbers is a clever use of the construction method and the relationship between an and sn!!

2sn=an^2+n-4

Subtracting 2s(n-1)=a (n-1)+n-1-4 yields: 2an=an 2-a (n-1)+1, i.e., (an-1) =a(n-1) an=a(n-1)+1, i.e. an-a(n-1)=1 is an equal difference series.

an=a1+(n-1)d=-3+n-1=n-4

s4=5s2

s4-s2=a4+a3=4s2

s2=a1+a2

a4=3q;a2=3 Cons state q; a1=3 q 2So:3(1+q)=4*(3 q+3 q 2)q 2=4q=-2

an=a3*(q)^(n-3)=2*(-2)^(n-3)=-2)^(n-2)

1）s3=7=a1+a2+a3

a1+3, 3a2， a(3)+4

That is: 6a2=a1+3+a3+4

6a2=7-a2+7

a2=2 and a1+a2+a3=2 q+2+2q=7q=or 2

q>1, so q=2

an=a2*q (n-2)=2*2 (n-2)=2 (n-1)2)bn ln a(3n+1)=ln2 (3n)=3nln2bn} is the difference series of the gods, b1=3ln2,d=3ln2tn=(b1+bn)n 2=3 2·n(n+1)ln21)n=1, a1=s1=1 3(an-1)a1=-1 rent-and-tear 2

a1+a2=s2=1/3(a2-1)

a2=1/4

sn＝1/3(an-1)

s[n-1]=1/3(a[n-1]-1)

an=sn-s[n-1]

3an=an-1-a[n-1]+1=an-a[n-1], i.e., 2an=-a[n-1].

an/a[n-1]=-1/2

So it is a proportional series, and the common ratio q=-1 2

an=a1*q^(n-1)=(1/2)^n