An algebra problem by math man
What is the sum of the first 100 non-negative integers??
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11 solutions
If you count 0 as the first number your frequency (n) is actually 101 meaning the right answer is 101/2(0+(101-1)1) = 101/2(100) = 5050.
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I think he is right if the frequency is 100 . Because the first 100 non negative integers are 0,1,2,....,97,98,99.. there are 100 integers. 100 is not included
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The first non-negative integers must be 1,2,3,...,98, 99, 100 'cuz 0 is neither negative nor positive, so 5050 is the answer. If I'm wrong then correct me
http://www.daviddarling.info/encyclopedia/G/Gauss.html This is the same question and same answer Gauss got himself. Read the seventh line of the first paragraph where it says "multiplying 101 by 50" which equals 5050.
If 100 wasn't included then the equation above would be the sum of the first 99 numbers which is 4950, if you add 100 to this result then we get the real sum of the 100 numbers which is 5050 because 100 is the last number that wasn't added.
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@Wesam Khalil – No. The question Gauss solved was sum of First 100 Natural Numbers . The Question asked here is some of first 100 whole numbers. ( Non-negative integers) .
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@Shashank Kancherla
–
People are over complicating this question. If we find the sum of 1 to 4 it would be no different from the sum of 0 to 4 because adding zero makes no difference.
1+2+3+4=10
0+1+2+3+4=10
and the reason people are making this mistake is because they're either using 0 as "a" which is pointless because you don't have to count 0 and you could start at 1, but if you do start from 0 then you also have to put the frequency (n) as 101 because there are that many numbers from 0 to 100.
0+1+2+3+4
1, 2, 3, 4, 5 there are five numbers even though the highest number is 4 because we are counting 0.
So if you count 0 as "a" then you have to count the frequency as 101, but if you make "a" 1 then your frequency is 100.
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@Wesam Khalil – No, If you are saying that , You are not considering zero as a non negative integer. It doesn't make any change to sum but , It is a number and has to be counted as a number. Ignore any formulae and what would you consider the sum of first two non negative Integers - 0+1 or 1+2 ? If you say it is the second, That means you do not know the significance of zero !
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@Shashank Kancherla – Oh! I see, I thought it was 0 to 100 but it was the first 100 non negative integers (0 to 99), thanks for the explanation.
How can you say that 0 is a positive integer? According to me, 0 is not a positive integer nor a negative integer. @Abhishek Chopra
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the question is asking for non-negative integers so it is saying all integers except the negative one, the question is not asking positive integers,
0 is not positive but neither it is negative , so it is included in our set then just use the formula of AP and get the answer
Someone needs to complain, about this post, within 1 to 100, there are 50 number bonds that add up to 101 and 50 x 101 = 5050. And I checked this, if you search this problem online you will get 5050.
we should not consider zero because "0" is neither an +ve integer nor -ve integer . so first non negative integer starts from "1"
n (n-1)/2...... n=100,....... 100 (100-1)/2=4950
0+(1+2+3+............+99) =0+99(99+1)/2 =4950
The first 100 non-negative integers :0,1,2,3,...,98,99. As first 100 are asked 100 cannot be counted as 0 is counted. The formula for The Sum of All Numbers till is [n(n+1)]/2. Therefore, [99(99+1)]/2 = [99(100)]/2=4950. Therefore 4950 is correct
49 pairs that sum to 100 or....1+99; 2+98,3+97....up to 49+51 and then plus 50 = 4950
The first non-negative integers start from 0 − 9 9
Use this formula
S n = Sum of a − U n
d = different
a = first number
n = sequence long
U n = n t h number
S n = 2 n ( 2 a + ( n − 1 ) d
Using U n = a + ( n − 1 ) d
S n = 2 n ( a + U n
for n = 1 0 0 , a = 0 , U 1 0 0 = 9 9 , d = U 2 − U 1 = 1 − 0 = 1
Subtitude:
S 1 0 0 = 2 1 0 0 ( 0 + 9 9 )
S 1 0 0 = 5 0 × 9 9 = 4 9 5 0
write the whole series forward like 0+1+2+3+...+99. then write it backwards 99+98+97+96+...+1+0 and add 'em up to get (99+99+99+...100times). but this is twice the sum of 0-99 so divide by 2 to get actual sum, i.e, 1/2x(99x100)=4950 . gauss style!!!!
First 100 non negative integers are 0, 1, 2........99. The formula for the sum of first "n" integers is: n(n+1)/2. Applying this formula the answer is: 99 (99+1)/2 =99*100/2 = 4950
first non-negative integers 0,1,2,...,97,98,99; we sum 1+99=100; 2+98=100; . . . 48+52=100; 49+51=100; and the rest 0 and 50; so, (49*100+0+50)=4950
this answer must be -4950 not 4950
please read it is non-negative, and not negative, even if it was written negative , then also answer would not come -4950 as 0 is then also not included .
Can you explain why the answer must be -4950?
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a=0 d=1 n=100 sum of 100 terms=100/2(2*0+99) =4950