High School Math Applications, High School Math Applications

N+1 is a corner mark, right?!

1）2(an+1-(n+1)+2)=an-n+2an+1-(n+1)+2/an-n+2=1/2a1-1+2=3

So it is a proportional series with a common ratio of 1 2 and a first term of 3.

an-n+2=1 to the n-1 power of 2.

bn=an-n+2=1 to the n-1 power of 2.

Let's start with the first question! The second question is to use the dislocation subtraction method, and the power or something will cause trouble

From the meaning of the title, an=n-1(n>=2), bn=n-1-n+2=1(n>=2)b1=2-1+2=3

an=(2+1+..n-1)n/2=(n+2)n/2(n>=2),a1=2bn=(3+1+..1)n/2=(n+2)n/2(n>2),b1=3

When n=1, a1+cb1=2+3c is OK.

When n>=2, an+cbn=(c+1)(n+2) 2When n=2, an+cbn=2(c+1).

When n=3, an+cbn=5 2(c+1).

When n=4, AN+CBN=3(C+1).

3(c+1)-5 2(c+1)=5 2(c+1)-2(c+1) can be arbitrary.

1) an=n-1(n>=2)；

b1=a1-1+2=3；

bn=n-1-n+2=1；

bn should be written separately.

2）an=2+n*(n-1)/2

bn=n+2

Let dn=(an+cbn) n=[4+n*(n-1)+2c*(n+2) 2n].

dn+1-dn=constant=n2+n-4-4cdn-dn-1=constant=n2-n-4c-4;

The two are equal.

n=0.So c doesn't exist.

no no no i don't no I'm an American I can't read Chinese.

Solution: (1) When x=0, t=0;When 00 function y=x+1 x increases monotonically.

y [2, in summary, t [0,1 2](2) when a [0,1 2], f(x)=g(t)=丨t-a丨+2a+2 3= 3a-t+2 3,0 t a; t+a+2/3,a≦t≦1/2.

g(0)=3a+2/3 g(1/2)=a+7/6 g(0)-g(1/2)=2a-1/2

m(a)=﹛g(1/2),0≦a≦1/4；g(0), 1 4 if and only if a 4 9, m(a) 2

A [0,4 9] does not exceed the standard, a (4 9,1] exceeds the standard.

(1) Total amount of projects: 20 24 60 = 28800 (vehicles·min).

2) The first 20 minutes to complete: 20 1, the second 20 minutes to complete 20 2,。。

The 25th 20 minutes to complete 20 25, so that 8 hours to complete: 20 1 + 20 2+. 20 25=20 (1+25) 25 2=6500 (vehicles·min).

There are 24-8 = 16 hours, which can be completed: 25 16 60 = 24000 (car·min).

6500 + 24000 = 30500 28800, can complete the task according to the requirements of the command.

The average 20-minute workload of each engineering vehicle is 1 (24*20*3)=a, a total of 3*24=72 20-minute in a day, but only 25 20-minute engineering vehicles are all driven.

Let the total amount of work be y

y=a*(72+71+70+..47)=a*(<72+47>*25 2) By calculator.

y is approximately equal to 》=1 so the task can be completed.

According to the conditions, the number of new license plate numbers added each year can be set to x, and the number of series An is the number of cars (10,000 units) in the nth year

The number of cars in the first year (2010) is A1 = 90A2 = A1 (1-6%) x = 90 +x (excluding scrap and adding the number of new license plates).

a3 = a2×(1-6%) x = 90× +x...

an = 90× +1+

According to the urban planning an 180 is true for all natural numbers n.

i.e. 90 +x 180

The right side of the x inequality is seen as a function of n, and the denominator t = (then the function is f(t) = = monotonically decreasing.

Thus the function on n decreases monotonically.

Therefore, the minimum value to the right of the inequality should be the limit where n tends to infinity. The limit value is so that the number of new license plates per year does not exceed 10,000.

Assuming that the previous year's ownership is y, and X license plate numbers are fixed every year, assuming that all license plate numbers can be distributed, then the current year's ownership is y=x+so x+<=180

x<=。

The solution is iterative and the initial condition y0=90

yi=。The objective function xi=

Accuracy|xi-x(i-1)|<

Result: 927835 (to the single digit).

By what time no more than 180 vehicles.

Obviously, this is a question of finding the minimum value.

The first thing you should think about is that there should be a stretch of road on the edge of AB, and then you should go straight from a point between AB to C.

The total distance is just like it says upstairs.

This method is the derivative.

To find the minimum value is to find its derivative, so that the derivative is equal to 0, and the resulting value is substituted for the required minimum value.

Solution: Understand the profit after one year first:

Cost: (100,000 + 10,000 * 1) * g (1) + 1 million * 1;Income: (100,000 + 10,000) * 100 = 11 million.

Profit: Revenue - Cost.

So: after n years, f(n) = (100,000 + 10,000 * n) * 100 - (100,000 + 10,000 * n) * 80 root number (n + 1) + 1 million * n).

1000-(10+n)*80 root number (n+1).

Find the maximum value of f(n), that is, find the minimum value of (10+n) root number (n+1);

Let y=root number(n+1), that is, find (9+y squared) y=9 y+y》=2*open((9 y)*y)=6

So, the highest profit = 1000-80 * 6 = 5.2 million, after the third year. （9/y=y==>y=3）

1) From 100 to 1000 y 40x·80 32x1000 to 5000 y 70 (x 1000) 40 1000 40 27x 24000

5000 to 10000 y 60 (x 5000) 40 5000 40 24x 80000

More than 10000kg y 50 (x 10000) 40 10000 40 20x 200000

To sum up, it is indicated by braces.