Q: What is the sum of the integers from 1 to ( and including ) 100?

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101

It is 100*(100+1)/2 = 50500.

It is 2500.

n(n+1)/2 5050

SUM = 0: FOR N = 1 to 100: SUM = SUM + N: Next N: PRINT CHR$(11); "Sum ="; SUM: END

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The sum of the integers from 1 to 100 inclusive is 5,050.

101

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

SUM = 0: FOR N = 1 to 100: SUM = SUM + N: Next N: PRINT CHR$(11); "Sum ="; SUM: END

Taking all the extreme numbers between 1 and 100 in each sum gives 1 + 100, 2 + 99, 3 + 98... and so on - fifty pairs of sums which add to 101. Therefore, the sum of the integers between 1 and 100 is equal to 50 x 101 = 5050.

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

First we find the sum of the integers from 2 to 100 including 2 and 100, then we divide by 99 since there are 99 numbers. There are 100 numbers between 1 and 100 and we are excluding only the number 1. If we want to exclude the number 2, and count only the numbers 3,4,5....100, this can be done with the same procedure and a slight modification. So the sum of the numbers 1 to 100 can be found by writing the numbers 1, 2,3,...100 Now write them backwards, starting at 100,99,98....1 Each column has a sum of 101 and there are 100 columns. So the total is 100x101, but we wrote the list twice so we must divide by 2 The sum is 100x101/2=50x101=5050 Now remember we want only 2 to 100 so the sum we seek is 5049. Since there are 99 number, the mean is 5049/99=51 In general the sum of the first n positive integers is n(n+1)/2 This can be proved the way we did or by induction.

The sum of integers from 1 to 2008 = 2008*2009/2 = 2017063