If the annual interest rate is 10%, what is the final value of 1000 yuan in five years? What is the present value? Please write the calculation formula, If the annual interest rate is 10%, what is the final value of 1000 yuan in five years? What is the present value? Please write down the calculation formula, I'll use it tomorrow,

1.1000(1+10%)^5=1000*1.61051=1610.51
2.1000/1.61051=620.92

What do you mean by the formula of compound interest terminal value and present value Final value formula F = PX (F / P, I, n) The present value formula P = FX (P / F, I, n) How to use the formula to calculate?

f: This is the formula for calculating the time value of capital. The corresponding coefficient can be found by looking up the coefficient table of compound interest present value and the coefficient table of final value of welfare. For example, if the annual interest rate is 10%, the cash of 10000 yuan is 10%

As for the calculation formula of the final value of ordinary annuity, it is still unclear how it came about The book says: F = a (1 + I) ^ 0 + a (1 + I) ^ 1 + a (1 + I) ^ 2 +... + a (1 + I) ^ n-2 + a (1 + I) ^ n-1 In my opinion, it is as follows: at the beginning of the first year, the deposit of a yuan is a (1 + I) at the end of the first year, which is not equal to a (1 + I) ^ 0 in the formula. At the end of the second year, the principal and interest of the first year is a (1 + I) plus a deposited in the second year as the principal of the second year. The final value of two years is [a (1 + I) + a] (1 + I), But this does not mean a (1 + I) ^ 0 + a (1 + I) ^ 1 in the formula?

In fact, this formula should be reversed. A (1 + I) ^ 0 refers to the last amount of money deposited, because ordinary annuities occur at the end of the period, which is also called post payment annuity. Therefore, the last amount of money has no interest
A (1 + I) ^ n-1 is the principal and interest of the money deposited at the end of the first year. Because it is deposited at the end of the first year, there is no interest in the first year. Therefore, the number of years in which interest is generated should be n-1 years

On the derivation of the final value formula of ordinary annuity

Suppose the terminal value is s, the annuity is a, the interest rate is I, and the number of periods is n: s
S=A+A（1+i）+…… +A（1+i）^n-1
Multiply both sides of the equation by 1 + I to obtain:
1+iS=A（1+i）+A（1+i）^2…… +A（1+i）^n
The former formula can be reduced
iS=A(1+i)^n-A
There are: S=A[(1+i) ^n-1]/i
In fact, this is the sum of an equal ratio sequence whose first term is a, the common ratio is (1 + I), and the number of terms is n
First term × (the nth power of 1-common ratio) × (1-common ratio)
It can be concluded

What are the formulas for calculating the terminal value and principal of series cash flow?

F = P × (1 + I) n (power) f: final value
P: Principal
i: Interest rate
N: Integral multiple of the time of interest rate acquisition

0

Subtract 1 + I with * 1 + I at the same time
The result is f = (a (1 + I) ^ n + 1 - A (1 + I)) / I
Instead of F = (a (1 + I) ^ n - A (1 + I)) / I

All the math formulas for Grades 1 to 6

1 number of copies per copy × number of copies = total number of copies △ number of copies = total number of copies △ number of copies = number of copies 21 times × multiple = several times several times △ 1 times = multiple several times △ times = 1 times 3 times speed × time = distance △ time = speed 4 unit price × quantity = total price △ unit price = total quantity △ quantity = unit price 5 Work efficiency × working time = total work, total work, work efficiency, work time, work time, work efficiency, 6 addend + addend, sum, one addend = another addend, 7, minus, subtraction, subtraction, subtraction, subtraction, subtraction, subtraction, subtraction, subtraction, 8 factors × factor, product, one factor, another factor, 9, divisor, divisor, quotient Divisor △ quotient = divisor quotient × divisor = divisor elementary school mathematics figure calculation formula 1 square C perimeter s area a side length perimeter = side length × 4 C = 4A area = side length × side length s = a × a 2 cube V: Volume A: edge length surface area = edge length × edge length × 6 s table = a × a × 6 volume = edge length × edge length × edge length v = a × a × a 3 rectangle C perimeter s area a side perimeter = (length + width) × 2 C = 2 (a + b) Area = length × width s = AB 4 cuboid V: Volume s: Area A: length B: width H: height (1) surface area (length × width + length × height + width × height) × 2 s = 2 (AB + ah + BH) (2) volume = length × width × height v = ABH 5 triangle s area a bottom h height area = bottom × height △ 2 s = ah △ 2 triangle height = area × 2 △ bottom triangle bottom = area × 2 △ height 6 parallelogram s area a bottom h height area = bottom × height s = ah 7 trapezoid s area A upper bottom B lower bottom h height area = (upper bottom + lower bottom) × height △ 2 s = (a + b) × h △ 28 circular s area C circumference Π d = diameter r = radius (1) circumference = diameter ×Π = 2 ×Π × radius C = Πd = 2 Π R (2) area = radius × radius ×Π9 cylinder V: Volume H: height s; Bottom area R: bottom radius C: bottom perimeter (1) side area = bottom perimeter × height (2) surface area = side area + bottom area × 2 (3) volume = bottom area × height (4) volume = side area △ 2 × radius 10 cone V: Volume H: height s; bottom area R: bottom radius volume = bottom area × height △ 3 total number △ total number = average sum difference formula (sum + difference) △ 2 = large number (sum difference) △ 2 = decimal number The problem of tree planting on an unclosed line can be divided into the following three situations: (1) if trees are to be planted at both ends of an unclosed line, Then: number of trees = number of sections + 1 = total length × (number of plants) - 1 total length = spacing × (number of plants-1) spacing = total length × (number of plants-1) if trees are to be planted at one end of the non closed line, then: number of trees = number of sections = total length × total length = spacing × number of trees = total length × (number of trees) × (3) if trees are not to be planted at both ends of non closed lines, Then: the number of trees = number of segments-1 = total length × (number of plants + 1) spacing of trees = total length × (number of trees + 1) 2 the relationship between the number of trees on the closed line is as follows: number of trees = number of sections = total length × total length of plant spacing = spacing × number of trees = total length × number of trees = total length × (profit + loss) × difference of distribution amount = number of shares participating in distribution (large profit - small profit) × difference of two distribution amount = number of shares participating in distribution (large loss - small loss) × difference of two distribution amount = number of shares participating in distribution meeting distance = speed and × meeting time meeting time = meeting distance △ speed and speed sum = meeting distance △ meeting time and problem tracing distance = speed difference × pursuit time Follow up time = pursuit distance △ velocity difference velocity difference = pursuit distance ／ pursuit time flow problem downstream velocity = still water velocity + water flow velocity countercurrent velocity = still water velocity + countercurrent velocity ＝ (downstream velocity + countercurrent velocity) × 2 flow velocity = (downstream velocity countercurrent velocity) × 2 concentration problem solute weight + solvent weight = solution weight Weight of solute ×weight of solution × 100% = weight of solution × concentration = weight of solute × weight of solute △ concentration = weight of solution, profit and discount problem, profit = selling price cost profit margin = profit × cost × 100% = (selling price ＋ cost-1) × 100% up and down amount = principal × percentage discount = actual selling price ＋ original selling price × 100% (discount ＜ 1) Interest = principal × interest rate × time, after tax interest = principal × interest rate × time × (1-20%)

Mathematical formulas from grade one to grade six

1. The number of copies per copy = the total number of copies △ the number of copies per copy = the total number of copies △ the number of copies = the number of copies per copy 2,1 times × multiple = several times several times △ 1 times = multiple several times 3. Speed × time = journey distance △ speed = time distance △ speed 4. Unit price × quantity = total price

Mathematical formula of grade six in primary school

1. The number of copies per copy = the total number of copies ／ the number of copies per copy = the total number of copies ／ the number of copies = 2, 1 times × multiple = several times, several times △ 1 times = several times, 3, speed × time = distance, distance, speed, speed, time, distance, time, speed, unit price, quantity, total price, total price, etc

Well, I can + + points!

Common mathematical formulas
Number of copies × number of copies = total number of copies / number of copies = total number of copies / number of copies = number of copies
1 times × multiple = several times several times △ 1 times = multiple several times × = 1 times
Speed × time = distance / distance / speed = time / distance / time = speed
Unit price × quantity = total price / unit price = total quantity / total price / quantity = unit price
Work efficiency × working time = total work, total work × work efficiency = working time
Total work × working time = work efficiency
Addend + addend = and - one addend = another
Subtraction minus = subtraction subtraction = subtraction difference + subtraction = subtraction
Factor × factor = product product △ one factor = another factor
Divisor △ divisor = quotient divisor △ quotient = divisor quotient × divisor = divisor
Primary school mathematics figure calculation formula
square
C perimeter s area a side length
Perimeter = side length × 4 C = 4A area = side length × side length s = a × a
cube
5: Volume a: edge length
The length of the edge is 6 × the length of the edge
V=a×a×a
rectangle
C perimeter s area a side length
Perimeter = (length + width) × 2 C = 2 (a + b)
Area = length × width s = ab
Cuboid
5: Volume s: Area A: length B: width H: height (1) surface area (length × width + length × height + width × height) × 2 s = 2 (AB + ah + BH)
(2) Volume = length × width × height v = ABH
triangle
S area a bottom h high area = bottom × height △ 2 s = ah △ 2
Height of triangle = area × 2 △ bottom of triangle = area × 2 △ height
parallelogram
S area a bottom h height
Area = bottom × height s = ah
trapezoid
S area a upper bottom B bottom h height
Area = (upper bottom + bottom) × height △ 2 s = (a + b) × h △ 2
circular
S area C circumference Π d = diameter r = radius
(1) Circumference = diameter ×Π = 2 ×Π × radius C = Πd = 2 Π R
(2) Area = radius x radius x Π
cylinder
v: Volume H: height s; bottom area R: bottom radius C: bottom perimeter
(1) Side area = perimeter of bottom surface × height (2) surface area = side area + bottom area × 2
(3) Volume = bottom area × height (4) volume = side area △ 2 × radius
Cone
v: Volume H: height s; bottom area R: bottom radius
Volume = bottom area × height △ 3 total number △ total number = average number
The formula of sum difference problem
(sum + difference) △ 2 = large number (sum difference) △ 2 = decimal
The problem of sum times
Sum × (multiple-1) = decimal decimal × multiple = large number (or sum decimal = large number)
Differential multiple problem
Difference × (multiple-1) = decimal number × multiple = large number (or decimal + difference = large number)
Tree planting
1. Tree planting on non closed lines can be divided into the following three situations:
(1) if trees are to be planted at both ends of an unclosed line, then:
Number of plants = number of segments + 1 = total length △ plant spacing-1
Total length = plant spacing × (number of plants - 1)
Plant spacing = total length × (number of plants - 1)
(2) if trees are to be planted at one end of an unclosed line and not at the other end, then:
Number of plants = number of segments = total length × plant spacing
Total length = spacing × number of trees
Plant spacing = total length △ number of trees
(3) if trees are not planted at both ends of the unclosed line, then:
Number of plants = number of segments-1 = total length × plant spacing-1
Total length = plant spacing × (number of plants + 1)
Plant spacing = total length × (number of plants + 1)
2 the quantitative relationship of tree planting on closed lines is as follows
Number of plants = number of segments = total length × plant spacing
Total length = spacing × number of trees
Plant spacing = total length △ number of trees
Profit and loss
(profit + loss) × the difference between the two distributions = the number of shares participating in the distribution
(large profit small profit) × the difference between the two distributions = the number of shares participating in the distribution
(large deficit small deficit) × the difference between the two distributions = the number of shares participating in the distribution
Encounter problem
Encounter distance = speed and X encounter time = encounter distance ＋ speed and speed sum = encounter distance ＋ encounter time
Follow up on Problems
Chase distance = speed difference × time pursuit and time pursuit and time = pursuit distance × speed difference speed difference = pursuit and distance ＋ pursuit and time
Flow problems
Downstream velocity = still water velocity + current velocity countercurrent velocity = still water velocity - current velocity
Still water velocity = (downstream velocity + countercurrent velocity) × 2 flow velocity = (downstream velocity - countercurrent velocity) × 2
Concentration problem
Weight of solute + weight of solvent = weight of solution weight of solute / weight of solution × 100% = concentration
Weight of solution × concentration = weight of solute △ concentration = weight of solution
Profit and discount
Profit = selling price - cost
Profit rate = profit ＋ cost × 100% = (selling price ＋ cost-1) × 100%
Up / down amount = principal × up / down percentage
Discount = actual price × original price × 100% (discount < 1)
Interest = principal × interest rate × time
After tax interest = principal × interest rate × time × (1-20%)
There are 5 cases of apples, each box has the same number. If 18 apples are taken out from each box, the remaining apples are exactly equal to the number of apples in the original three boxes. How many apples are there in each box
The total number of apples taken out: 18 × 5 = 90 (pieces), because the remaining apples are exactly equal to the number of the original 3 boxes of apples, and the number of apples taken out is exactly 2 (5-3) boxes. Therefore, we can get that there are apples in each box: 90 ／ 2 = 45 (apples)