The sum of all even numbers between 0 and 398 (2-396) is 157,212. If you include 398 the sum is 158,802.
500
10 + 0 =10 ?
The sum of all numbers between 1 and n = n*(n+1)/2 The sum of all numbers between 1 and 999 = 999*1000/2 = 499500 The sum of all numbers between 1 and 9999 = 9999*10000/2 = 49995000 So, the sum of all numbers between 1000 and 9999 = 49495500
You know that sum of the first n whole numbers is n(n+1)/2. ( it is the same as the first n natural numbers since the zero does not add anything) So lets say you want the sum of all the whole numbers between 3 and 10. ( I made it easy to illustrate the idea.) The sum of the whole numbers between 0 and 3 is 3(4)/2=6 The sum of the whole numbers between 0 and 10 is 10(11)/2=55 So the sum of the whole numbers between 3 and 10 is the (sum of the whole number between 0 and 10) -(sum of whole numbers between 0 and 3) which is 55-6=49 So in general, for whole numbers m and n with m
The sum of all even numbers between 0 and 398 (2-396) is 157,212. If you include 398 the sum is 158,802.
500
55
If this question means "in the interval 0 to 16 inclusive, is the sum of the odd numbers the same as the sum of the even numbers ?" then the answer is no. The sum of the even numbers is eight more than the sum of the odd ones.
10 + 0 =10 ?
The sum of all numbers between 1 and n = n*(n+1)/2 The sum of all numbers between 1 and 999 = 999*1000/2 = 499500 The sum of all numbers between 1 and 9999 = 9999*10000/2 = 49995000 So, the sum of all numbers between 1000 and 9999 = 49495500
You know that sum of the first n whole numbers is n(n+1)/2. ( it is the same as the first n natural numbers since the zero does not add anything) So lets say you want the sum of all the whole numbers between 3 and 10. ( I made it easy to illustrate the idea.) The sum of the whole numbers between 0 and 3 is 3(4)/2=6 The sum of the whole numbers between 0 and 10 is 10(11)/2=55 So the sum of the whole numbers between 3 and 10 is the (sum of the whole number between 0 and 10) -(sum of whole numbers between 0 and 3) which is 55-6=49 So in general, for whole numbers m and n with m
to print the sum of first ten numbers:- void main() {int i,sum; sum=0; while(i<=10) sum=sum+i; i=i+1; } printf("The sum is : %d",sum); }
The sum of the oxidation numbers of all carbons in C6H5CHO is equal to 0. This is because the oxidation number of carbon is -2 in CH3CHO, and there are six carbons in C6H5CHO. So, 6 * (-2) = -12. However, since the overall charge of the molecule is 0, the sum of the oxidation numbers of all carbons must be 0 to balance the charge.
For the product to be zero, one of the numbers must be 0. So the question is to find the maximum sum for fifteen consecutive whole numbers, INCLUDING 0. This is clearly achived by the numbers 0 to 14 (inclusive), whose sum is 105.
It is 180,001.
You add the numbers in a loop. Here is an example in Java:int sum = 0;for (int i = 1; i