n(n+1)/2
5050
101
It is 100*(100+1)/2 = 50500.
It is 2500.
SUM = 0: FOR N = 1 to 100: SUM = SUM + N: Next N: PRINT CHR$(11); "Sum ="; SUM: END
It's easy to work it out yourself.... Multiply 100 by 49, add 50, add 100 - and you have your answer !
The sum of the integers from 1 to 100 inclusive is 5,050.
101
It is 100*(100+1)/2 = 50500.
It is 2500.
The following will sum all integers from 1-100. Adjust according to your needs.int sum = 0;for (int i = 1; i
SUM = 0: FOR N = 1 to 100: SUM = SUM + N: Next N: PRINT CHR$(11); "Sum ="; SUM: END
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.
Write a program to find the number and sum of all integers from 100 to 300 that are divisible by 11
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 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