Write a program to calculate the sum of the natural numbers one plus two plus three up to 100. Two s

5 Ways to Speak C

while statement.

#include ""

void main()

int x,i;

x=0;i=1;

while(i<=100)

x+=i;i++;

printf("1+2+……100=%d",x);

do...while statement.

#include ""

void main()

int x,i;

x=0;i=1;

dox=x+i;

i=i+1;

while(i<=100);

printf("1+2+……100=%d",x);

for statement.

#include ""

void main()

int i,sum=0;

for(i=1;i<=100;i++)

sum+=i;

printf("1+2+……100=%d",sum);

goto statement and do....while equivalent.

#include ""

void main()

int i=1,sum=0;

loop:sum+=i;

i++;if(i<=100)goto loop;

printf("1+2+……100=%d",sum);

The goto statement is equivalent to while.

#include ""

void main()

int i=1,sum=0;

loop:if(i<=100)

sum+=i;

i++;goto loop;

printf("1+2+……100=%d",sum);

Is it available in math books for high school students.

Of course it's 0, like the square root of 1 + 1-1 two add to 0, and although there are infinite natural numbers, their square roots have plus or minus two, which add up to 0

0 points, who wants to tell you, it's really stingy, and I want others to answer questions with you and don't want to give people points.

No one likes to do it for you with a reward of 0 points, at least 20 points, I can still write about it, it's quite simple.

Oh, here's the thing, add 1 and 99. 2 and 98 are added, 3 and 97 want to be added... 49 and 51. It's exactly 50 100. It's 5000, and there's one 50 left that can't be combined, so it's 5050 in total

1+2+3+……100

This is the summation of the series of equal differences, and there is a summation formula as follows:

Let the difference series a1, a2, a3 ......An has a total of n terms, the first term is A1, and the tolerance d=A2-A1=A3-A2=......=an-a(n-1), then s=a1+a2+......an, there is:

s=(a1+an)*n 2=(first term + last term)*number of terms 2 Also, an=a1+(n-1)d, that is, if you only know the first term a1 and the number of terms n, the tolerance d

It can also be summed by the transformation of the formula, which is as follows:

s=(a1+an)n/2=[a1+a1+(n-1)d]n/2=n[a1+(n-1)d/2]

The number of first and final multipliers divided by 2 (1+100)*100 2=5050

There is no mathematical formula, the series diverges, but there is an estimating expression, and it is close to the natural logarithmic difference to the Euclidean constant.

It's not necessary, use a computer to calculate, a layer of loops, a built-in counter and an accumulator can easily be done

Not bad, not 2,1+1 2+1 4+1 8+...1/(2^n);(n is infinite) is 2, misremembered.

Calculate it with vb, the source ** is here:

private sub form_load()dim i as integer,result as doublefor i=1 to 100

result=result+1/i

nextmsgbox str(result)

end sub

Give the result of the run:

See! Such a result, even if it is calculated by hand, is a large number, and there is no need to calculate it by hand.

It's very easy, and it can be solved with the knowledge of the first year of high school, that is"Sum of the sequence"

sn＝1 + 1/2 + 1/3 + 1/4 + 1／100sn＝1＋（1－1／2）＋（1－2／3）＋（1－3／4）＋…1－ 99／100）

All the 1s are added to 100, and the formula for the remaining number series is an n (n+1), which can be split and eliminated.

luxiao1990 you are wrong! 1 2+1 3+1 6=1 What about others??

I thought it was about 2 at first! It's an illusion!!

The decimal system is not special.

So so pour out the numbers and.

It's not a perfect number.

Write a small program in C and you're good to go.

The crooked Ai Ziming on the second floor must be wrong.

I'm thinking about it.

1-100 sum of all numbers = 1+2+.100=(1+100)*100/2=5050

All even numbers sum = 2 + 4 + ...100=2(1+2+..50)=2(1+50)*50/2=2550

All odd sum=5050-2550=2500

The sum of all composite numbers = 5050 - 1 - 1060 = 3989

The first plus the last, the second plus the second-to-last, and so on, the resulting sum is equal, multiply the resulting sum by the number of data and divide by 2

(1 100)*100*(1 2) 5050 (the sum of the last items of the previous item is multiplied by the total number divided by 2; Available for even, odd, and both.

The sum of the square roots of all natural numbers from 1 to 100 is equal to 0

The square root of 1 is plus or minus 1Then plus 1 + minus 1 = 0

The square root of 10 is the positive and negative root number 10Then the positive root number.

10 + negative root number 10 = 0

So. The sum of the square roots of the other natural numbers from 1 to 100 is equal to 01, and the square root is plus or minus 1, which adds up to 0

The same is true for all numbers.

0 ? The square root of 1 is plus or minus 1 and adds up to 0

The same is true for all numbers.

Irrational numbers can't be opened, so write them directly.

Just card out the full square number.