Two simple high school math problems, number series!!!!!!!!!!!

1 Note In question 1, an2 denotes the second power of an.

2 indicates the 2nd power in parentheses.

is the multiplication sign n2 to the 2nd power of n.

an - 2/an)=2n

an2 - 2 = 2n*an

an2 -2 + n2 = 2n*an + n2 shift.

an2 + 2n*an + n2 = 2 + n2 is preceded by 2 secondary terms, so.

an + n)2 = n2 + 2

An + n = (n2 + 2) is rounded off because an<0 or an + n=- (n2 + 2).

To sum up: an=-n - n2 + 2).

2 a1+a2+..a98+a99=99

a1+a4+..a97= ..Equation 1

a2+a5+..a98= ..Equation 2

a3+a6+..a99= ..Equation 3

Obviously, Equation 1 is 33 smaller than Equation 2 (a2 - a1 = 1, a5 - a4 = 1 .99 numbers are divided into 3 groups, a group of 33, so.

Equation 2 is 33 bigger than 1 and Equation 3 is 33 bigger than 2

a1+a2+..a98 + a99 = Equation 1 + Equation 2 + Equation 3 = Equation 1 + Equation 1 + 33 + Equation 2 + 33 = Equation 1 + Equation 1 + Equation 1 = 99

So Equation 1 = 0

Equation 2 = 33

Equation 3 = 66

ok,.

Sorry, the first question will not.

Question 2: From the question, it can be seen that the difference series d=1

From the formula, it can be seen that s99=99

According to sn=na1+n(n-1)d 2, a1= and then sn= are calculated

The requirements can be reduced to s33+2+4+6+...66

s33+33(2+66)/2

I don't have a computer, so you're going to do it yourself!

I remind you to put down, 8To satisfy the increasing sequence, you only need to satisfy a(n+8)-a(n)>1, and then find the range 8 by yourselfProof 8a8=-8 8a8-..

In the same way, find a8 and then a8 a8 is not equal to a number and prove it, or use the counterproof method, assuming that it is an equal ratio series and then prove that there is no equal ratio also.

1.Let's solve the equation directly... Treat an as an unknown and solve n-root number under n-root +2

2.Calculate a1 first, and then know that there are 27 terms in the next one, and write each term in the form a1+nd, and since d is 1, then the above equation can be written.

27a1+2+5+..98 can be solved to -189

1。(1) Solution: The original equation is: an 2-2nan-2=0 from the root finding formula: an=n-under the root number (n 2+2).

2.It can be seen that a6-a3=3, a9-a6=3,..So, the original = a1 a2....a98＋a99-66=99-66=33

5 years = 60 months.

Tolerance d=2 yuan.

a1=2(1-1)+500=500

sn=a1*n+n(n-1)d/2

s60 = 500 * 60 + 60 * 59 * 2 2 = 33540 yuan total consumption 500000 + 33540 = 533540 yuan per day consumption:

533540 5 365 yuan.

Comparing n=k and n=k+1,2(k+1-1)+500-(2(k-1)+500)=2, it is obvious that the difference series.

January 500, February 502, March 504....Total 5*12=60 months.

Total cost = 50 * 10,000 + 500 * (12 * 5) + (0 + 118) * 60 2 = 533,540 yuan.

Cost per day: 533540 (365*5) = RMB.

The monthly cost is cumulative, with the difference series 2n+498, a total of 5*12 months.

The total consumption is a cumulative monthly expenditure of +500,000 yuan.

Average consumption per day = total consumption Total number of days.

1 Because it is an odd function, f(0)=0, b=0

Substituting f(1)=1 2, we get 1 (1+a)=1 2, a=12 f(x)=x (x 2+1).

f(x)'=[(x^2+1)-2x^2]/(x^2+1)^2=(1-x^2)/(x^2+1)^2

Req f(x).'>0 gives x to belong to (-1,1);

f(x) is an increasing function over the interval (-1,1).

3 g(0)=3^0-0=1

g(1)=1/3-1/2=-1/6<0

Whereas g(x) is continuous, so there is a zero point on (0,1).

So the function g(x) has a zero point on (-.

1.The inverse negative proposition is: x is not equal to 0 or y is not equal to 0, then xy is not equal to 0Wrong!

2.The negative proposition is: if it is not a square, it is not a diamond. Wrong!

3.The inverse proposition is: if a>b, then ac2 bc2. Wrong!

4.x2-2x+m = x-1)2+m-1>m-1Because m>2, x2-2x+m = x-1)2+m-1>m-1>0 correct!

The above propositions are true: 4

1.If xy=0, then x=0 and y=0 are the inverse of x=0 if x=0 or y=0, then xy=0That's right.

2.The square is a diamond-shaped proposition that is.

A square is not a diamond. Mistake.

3.If ac bc , then the inverse proposition of a b is .

If a b, then ac > bc. Mistake. If c=0 is not true.

4.If m 2, then the solution set of the inequality x -2x + m 0 is r.

Solution: x -2x + m 0

i.e. x -2x+1 1-m

i.e. (x-1) >1-m

If m>2, then 1-m<-1

The inequality is constant. The solution set is r. That's right.

To sum up, 1 and 4 are correct.

Is this a quiz question or a multiple-choice question?

Let's talk about the idea: Establish the function f(x)=1 Sidka 5*(100-x)+2 (root number x).

Replacing it with the root number x with t = is a quadratic function hole early, and I have learned Nalufinch in junior high school. Find f(x)max=

I don't understand hi me.