The sum of integers from 1 to 2008 = 2008*2009/2 = 2017063
2550
x+x+1=1002x=99x=49.5x+1=50.549.5+50.5=100.0
55
To calculate the sum of the numbers 1 to n, the formula is: sum = n(1 + n) / 2 So, an equation to find the sum of the integers 1 to 2010 is: sum = 2010 x (1 + 2010) / 2
It is 100*(100+1)/2 = 50500.
It is 2500.
n(n+1)/2 5050
The following will sum all integers from 1-100. Adjust according to your needs.int sum = 0;for (int i = 1; i
To find the sum of the first 100 integers, you first add 1 plus 100 (the first and last numbers of the set) and get 101. Do the same with the next two integers, 2 and 99 and you'll get 101. Since you are adding two integers at a time and there are 100 integers between 1 and 100, you'll get 101, fifty times. Therefore, a shortcut would be to simply multiply 101 times 50 and get 5,050.
The sum of integers from 1 to 2008 = 2008*2009/2 = 2017063
The sum of all integers from 1 to 20 inclusive is 210.
It's easy to work it out yourself.... Multiply 100 by 49, add 50, add 100 - and you have your answer !
The sum of the integers from 1 through 300 is 44,850.
The flowchart to read 10 positive integers K>10 Start A N K=1 Sum = 0 Sum = Sum + K2 B Is Y Print K > 100? sum K=k+1 End B A
The integers are 99, 100 and 101. There is also a set of consecutive even integers whose sum is 300. That set is 98, 100 and 102.
The sum of all the digits of all the positive integers that are less than 100 is 4,950.