Q: Sum the integers from 1 to 100?

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101

The following will sum all integers from 1-100. Adjust according to your needs.int sum = 0;for (int i = 1; i

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

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.

The sum of the integers from 1 through 300 is 44,850.

Write a program to find the number and sum of all integers from 100 to 300 that are divisible by 11

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.

No. The sum of all integers between 1 and 500 is 124,749.

It's easy to work it out yourself.... Multiply 100 by 49, add 50, add 100 - and you have your answer !

They are 2n+2

The sum of all the digits of all the positive integers that are less than 100 is 4,950.

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

10100.

203

2550

(300 x (300 + 1)) / 2 = 45150 Therefore, the sum of all the integers from 1 to 300 is 45150.

The sum of the first 201 positive integers is 20100 if you include 0 (i.e. from 0 to 200). If you sum the integers from 1 to 201 instead, the sum is 20301.

Usually all the integers (counting numbers) from 1 to 100.Usually all the integers (counting numbers) from 1 to 100.Usually all the integers (counting numbers) from 1 to 100.Usually all the integers (counting numbers) from 1 to 100.

The sum of the first 40 positive integers (1-40) is: 820

The first odd positive integers are "1" and "3" which the sum is 4.