Help solve 2 Olympiad problems!! Easy calculation, thank you very much, it s better to have an expla

It should be written as 1+1 (1+2)+1 (1+2+3)+1/（1+2+3+..100）

1+1/3+1/6+1/10+..1 n(n+1) 2 (1 plus one-third plus one-sixth.)

2*[1/2+1/6+1/12+1/20+..1 n（n+1)] The denominator of each term is multiplied by 2).

2*[1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+..1 n-1 (n+1)] Split each term, e.g. 1 6 = 1 2-1 3

2*[1-1/（n+1)]

2*n (n+1) is substituted for the value of n.

The second question should be written as (1-1 4)*(1-1 9)*(1-1 14)*1-1/100）

Each term can be seen as the product of the following two numbers: n-1) divided by n * n+1) divided by n

For example, 1-1 4=1 2*3 2

Original question. 1/2*3/2*2/3*4/3*3/4*5/4*..n-1)/1*(n+1)/n

1/2*(n+1)/n

n+1)/2n

Substitute the value of n.

Eh: Let's start by saying that half is 1 2 instead of...

1/（1+2+3+..n)=2 (n*(n+1))=2*(1 n-1 n+1) by 1+2+3+.n=n*(n+1) 2.

So the first formula = 2*(1-1 2+1 2-1 3+.1/100-1/101)=200/101

a2-b2=(a+b)*(a-b) So:

The second formula = (1 2) * (3 2) * (2 3) * (4 3) *99 100) * (101 100) = (1 * 2) * (101 100) = 101 200 (the first and last items remain).

I don't understand your question either, if 1+(2+1)+(1+2+3)+.Such a formula,"/1"Isn't it superfluous?

Now for you to sum.

Each item can be written in the form of a general term.

a1=1=1(1+1)/2

a2=1+2=2（1+2)/2

a3=1+2+3=3（1+3)/2

an=1+2+3+..n=n(n+1)/2a1+a2+a3+..an

1(1+1)+2(2+1)+3(3+1)+.n(n+1)]/2[1^2+2^2+3^2+..n^2+(1+2+3+..

n)] 2 to this step, the focus is to find the sum of squares, which is 1 2 + 2 2 + 3 2 + ...n 2 = n(n + 1) (2n + 1) 6 is substituted into the above formula.

n(n+1)(n+2)/6

Substitute the value of n.

Is that a division or a fraction?!?

(1) The second interception of the full length of 3 4x1 2=3 8, then the second interception of the total length of 1 4 + 3 8 = 5 8

2) The first time to use 30x1 5 = 6 tons.

The second time 6x5 4 = 15 2 tons were used.

Shared to 6 + 15 2 = tons (one of thirteen and two molecules) (3) left after the first day of transportation.

The next day there is 3 4-1 4x3 5 = 12 20 = 3 5 then the total cargo is 90 (3 5) = 150 tons.

4) Let his speed be 2x

The thief's speed is x

The speed of the car is (1+4 5)2x.

After 10 seconds, 2 people are separated from each other (x+(1+4 5)2x)*10=46x set the time t to catch up with the thief

46x+x*t=2x*t

Solution t = 46 seconds.

SOSO has the answer for you!

A tube, the first time to cut off a quarter of the total length, the second time to cut off the remaining half. How many fractions of the total length are truncated twice?

A quarter plus (one minus a quarter) multiplied by one half = five eighths.

A pile of yellow sand 30 tons, the first time used a fifth of the total, the second time.

One and a quarter times the first one, and how many tons of the second one?

30 * (1 + one and a quarter) * 5th = tons.

There is a batch of goods, a quarter of this shipment was transported on the first day, and the second day was three-fifths of the first day, and there are 90 tons left undelivered, how many tons is this shipment?

90 (1-quarter-quarter* three-fifth) = 150 tons.

One question and one question: 1. Set the construction period x days, then the efficacy of A is 1 (x-1), B is 1 (x+3), according to the conditions there is equation 1 (x-1) + x (x+3) = 1, the solution is x=3, then the efficacy of A is 1 2, B 1 6, and 2 3, so it takes days.

2. Set the construction period x days, then the efficiency of A is 1 x, and B is 1 (x+, according to the conditions there is an equation, the solution is x=77 13

3. Let the original x grams, from the salt quality does not become, the solution is x = 1200

4.Let the original x grams, from the water quality does not become, the solution is x = 1500

5. The original plan is x hours, x=, x=6

6. The original plan is x hours, 1200 x = 1200, x = 100

The day is one-fifth of the time, so A is 25 days and B is 30 days.

8. A is 20% higher than B is a multiple of B, so B time = 24 times of 20th, and the efficiency of the two is 11/120, so the cooperation time is 120ths of 11th.

9.B has twice as many people, so it's 30 and 45.

10.It takes 2 days to complete 40% A, so it takes 5 days for A alone, B efficiency = 1 4-1 5 = 1 20, and B alone takes 20 days.

11.Cooperation for 8 days is 2 5 of the project, so B alone 4 days did 2 15, so B alone 30 days, A efficiency = 1 20-1 30 = 1 60, A alone 60 days, efficiency 2:1

The solution is x=84, y=112, and it takes 56 days to calculate B.

13.For a task, the master and apprentice worked together for 2 days to complete three-fifths of all the tasks, and then the master stopped working for 2 days and continued to cooperate with the apprentice, and the work efficiency ratio of the master and apprentice was known to be 2:1.

Q: How many days did it take to complete the task? Three-fifths of the master's two-fifths apprentice one-fifth, so the master alone takes 5 days and the apprentice alone takes 10 days, and the remaining task is two-fifths, the apprentice does one-fifth in two days, and the last one-fifth cooperates with Yongle two-thirds of the day, so a total of 4 and two-thirds of the days.

Divide the situation, and then look at the changes in the ailment.

1) Take out 2 white balls.

The black ball in the first is unchanged, and the white ball is minus 1

2) Take out 2 black balls.

The black ball in the first is minus 2, and the white ball is added by 1

3) Take out 1 black and 1 white.

The black ball in the first is unchanged, and the white ball is minus 1

To sum it up, for each 1 take, the total number of A is reduced by 1

The number of black balls remains unchanged or minus 2, and the total number is always an odd number.

So the last one left must be the black ball.

Need to take: 90 + 91 - 1 = 180 times.

First make sure the last ball is black! 90 times. I want to score points in the process, thank you!!

Question 1: Choose C

If the numerator of this number must have the least common multiple of the denominator of these three numbers, and the denominator is the three greatest common factors. This number is 70 3, which is 23 and 1 3

Question 2: 1. You can draw 1, 2, 4, 7, 10, 13, 16, 19, 22, 25 ,......See the pattern, right? From 2 onwards, there are 670 multiples of 3 more than 1, but only 669 multiples of 1 more than 3 (2011 is no longer in the range of cards), plus 1 and 2, this draw can draw a total of 671 cards.

2. It can draw 1, 3, 5, 7, 9, 11 ,......See the pattern, right? A total of 1,005 tickets can be drawn in this method.

3. You can draw 2, 3, 4, 8, 9, 13, 14 ,......This method draws less than half the number of sheets drawn in 2010.

4. It can draw 2, 3, 6, 7, 11, 12, 16, 17 ,......The number of sheets drawn by this method is less than half that of 2010.

From the above analysis, it can be seen that only a maximum of 1005 tickets can be drawn.

Question 3: If x yuan is used to purchase zongzi, there are:

x=2500 yuan.

Answer; It costs 2,500 yuan to buy zongzi.

1.Let this number be m n (m,n coprime).

Yes: 3m (7n), 9m (5n), 3m (2n) are all natural numbers m is divisible by 7, 5, 2, m, m 7 5 2 = 703 n, 9 n is a natural number, n 3

m n 70 3 = twenty-three and one-third.

2.You can choose 1, 3, 5, 7 ,..There are a total of 1005 cards in 2009 (because all numbers are odd numbers, and the sum of any two odd numbers is even, it is impossible to have a card number equal to the sum of the numbers of the other 2 cards).

So, at least 1005 tickets can be selected.

It is proved below that 1006 sheets and above are impossible.

If 1006 cards are selected, the largest card number is n (n 2010), then the difference between n and the other 1005 card numbers is greater than or equal to 1, less than 2010 natural numbers, and not equal to each other, a total of 1005 natural numbers, the original selection of 1006 card numbers is also greater than or equal to 1, less than or equal to 2010, and not equal to each other, a total of 1006 natural numbers, 1005 + 1006 total 2011 natural numbers, These 2011 natural numbers are greater than or equal to 1 and less than or equal to 2010, then at least 2 are equal and contradictory.

So you can only choose a maximum of 1005 tickets.

3.It took x yuan to buy zongzi.

Yes: x=2500

It costs 2,500 yuan to buy zongzi.

Question 1: A certain number must satisfy the distinction and the seventh.

Three, fifths.

9. The multiplication of three-thirds of the total is a natural number, and this number is required to be as small as possible. Then the numerator of this number must have the least common multiple of the denominator of these three numbers, and the denominator is the greatest common factor of these three numbers. That number is 70 3, which is twenty-three and one-third.

Question 2: The question is not clear, is "if a number of cards are selected" random or customized?

Otherwise, I chose all 2010 sheets, because 1 is the smallest number, and the sum of any two numbers is greater than it.

Question 3: Set up to buy zongzi for x yuan.

Desired sales, the total sales of zongzi is, and the desired profit is a=

In the actual sales situation, the total number of purchases can be regarded as 1 copy.

Sales before the Dragon Boat Festival were.

After the Dragon Boat Festival, the sales volume is b=(the actual profit is 15% less than the desired profit, so (a-b) a=15%

Get x=2500