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
Oh, dude, you're hitting me with the math questions. So, when you add 1 plus 2 plus 3 all the way up to 100, you get... drum roll, please... 5050. Yeah, like, someone actually sat down and figured that out. Crazy, right?
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 !