Although the topic is very long, I only ask the first detail, In order to promote sales, an automobile sales company has adopted a more flexible payment method. On the premise that the payment for a 100000 yuan car is paid in full within one year, it can choose the following two different payment schemes to purchase a car: 1. Payment in three installments, the first payment four months after purchase, the second payment four months after purchase, and the third payment four months after purchase 2. Pay in 12 installments, the first one month after purchase, and the second one month after purchase, The monthly interest rate is 0.8%, and the monthly interest rate is calculated by compound interest. Which of the above two schemes has less total payment? A: for scheme 1, if the amount of each payment is x 10000 yuan, then after four months, the principal and interest of the first payment will be x times 1.008 ^ 80000 yuan, For scheme 2, the first time is x times 1.008 ^ 11, the principle seems to be the same as above, but why

Although the topic is very long, I only ask the first detail, In order to promote sales, an automobile sales company has adopted a more flexible payment method. On the premise that the payment for a 100000 yuan car is paid in full within one year, it can choose the following two different payment schemes to purchase a car: 1. Payment in three installments, the first payment four months after purchase, the second payment four months after purchase, and the third payment four months after purchase 2. Pay in 12 installments, the first one month after purchase, and the second one month after purchase, The monthly interest rate is 0.8%, and the monthly interest rate is calculated by compound interest. Which of the above two schemes has less total payment? A: for scheme 1, if the amount of each payment is x 10000 yuan, then after four months, the principal and interest of the first payment will be x times 1.008 ^ 80000 yuan, For scheme 2, the first time is x times 1.008 ^ 11, the principle seems to be the same as above, but why

Application of mathematical function in grade one of senior high school
The topic is as follows: a car company has adopted a more flexible payment method for sales promotion. On the premise that the payment for a 100000 yuan car is paid in full within one year, it can choose the following two installment payment schemes: scheme 1: pay in three installments, the first payment four months after the purchase, and the second payment four months later, The second plan is to pay in 12 installments, the first one month after the purchase, the second one month after the purchase, and the twelfth one month after the purchase. It is stipulated that the amount of each installment is the same, the monthly interest rate is 0.8%, and the monthly interest is calculated by compound interest, It means that the interest of last month should be recorded in the principal of next month. Compare the above two schemes, which has less total payment? Give the reasoning process. (reference data: 1.008 ^ 3 ≈ 1.024,1.008 ^ 4 ≈ 1.032,1.008 ^ 11 ≈ 1.092,1.008 ^ 12 ≈ 1.1)
 
 
 
 
 
Suppose that x ten thousand yuan should be paid each time
For option 1:
When the first payment (i.e. 12 / 3 months after the loan) is x 10000 yuan, the difference is (12-12 / 3 = 8) months when the payment is made,
The deposit period of this payment x 10000 yuan is 8 months, according to the compound interest formula
The sum of this payment and interest is: X (1 + 0.008) ^ (12-12 / 3) = x × 1.008 ^ 8
 
In the same way
For option 2:
When the first payment (i.e. 12-11 = 1 month after the loan) is x 10000 yuan, the difference is (12-1 = 11) months when the payment is made,
The deposit period of this payment x 10000 yuan is 11 months, according to the compound interest formula
The sum of this payment and interest is: X (1 + 0.008) ^ (12-1) = x × 1.008 ^ 11
 
 
Attached:
Using the knowledge of sequence, there is the formula of installment payment: x = a (1 + P) ^ m [(1 + P) ^ m / N - 1] / [(1 + P) ^ m - 1]
Related issues of installment payment

1. In the triangle ABC, 2sinacosb = sinc is known, so why is this triangle isosceles triangle?
2. The three sides of the acute triangle are 1.3 a, and the value range of a is?
3. In the triangle ABC, BC = a, AC = B, a, B are the two roots of the equation x ＾ 2-2 root sign 3 multiplied by X + 2 = 0, and 2cos (a + b) = 1
Find: length of angle c? AB?
4. In the triangle ABC, a = 3 times the root 3, C = 2, B = 150 ', then B =?

(1)
2sinAcosB=sinC
2cosB=sinC/sinA
2*(a^2+c^2-b^2)/2ac=c/a
a^2+c^2-b^2=c^2
a^2=b^2
a=b
A triangle is an isosceles triangle
(2)
The three sides of an acute triangle are 1, 3 and a respectively
Then a3-1
20,cosC>0
1^2+3^2>a^2,a^2+1^2>3^2,a^2+3^2>1^2
A root 8
So the value range of a is (root 8, root 10)
(3)
2cos（A＋B）＝1
cos(180-C)=1/2
cosC=-1/2
C=120
a. B is the two roots of the equation x ＾ 2-2 root sign 3 multiplied by X + 2 = 0
AB = 2, a + B = 2, radical 3
a^2+b^2=(a+b)^2-2ab=12-2*2=8
cosC=(a^2+b^2-c^2)/2ab=(8-c^2)/2*2=-1/2
c^2=10
C = root 10
AB = root 10
(4)
CoSb = (a ^ 2 + C ^ 2-B ^ 2) / 2Ac = cos150 = - radical 3 / 2
(27 + 4-b ^ 2) / (2 * 3 radical 3 * 2) = - radical 3 / 2
31-b^2=6
b^2=25
b=5

The function f (x) = 2 ^ X-2 ^ - x is known. The sequence {an} satisfies f (log2 an) = - 2n
1. Find the general term formula of sequence {an}. 2. Construct a new sequence {BN} by BN = an + N, and prove that {BN / N} is a decreasing sequence
The first one asked me how to calculate an-1 / an = - 2n, as if there were two sequences,

(1) After you calculate an-1 / an = - 2n, you can convert it to an-1 / an + 2n = 0
Then the two sides of the equation multiply an, that is, an ^ 2-1 + 2nan = 0, that is, an ^ 2 + 2nan-1 = 0
Then take an as an unknown number and use the formula of quadratic equation of one variable to find the root: x = (- B ± √ B ^ 2-4ac) / 2A
an=(-2n±√(2n)^2+4)/2=-n±√(n^2+1)
And because the sequence {an} satisfies f (log2, an) = - 2n
So an > 0
So an = √ (n ^ 2 + 1) - n
(2)bn=an+n=√(n^2+1)
bn/n=√(n^2+1)/n
Suppose N10
So {BN / N} is a decreasing sequence

Find the first item and tolerance
(1) An (where n is the lower subscript) = 2n + 7
(2) An (this n is the bottom subscript) = √ 2-2n

(1)a1=2+7=9
a2=4+7=11
d=a2-a1=11-9=2
(2)a1=√2-2
a2=√2-4
d=a2-a1=(√2-4)-(√2-2)=-2