The sum of the integers from 1 to 100 can be calculated using the formula for the sum of an arithmetic series: ( S_n = \frac{n(n + 1)}{2} ), where ( n ) is the last integer in the series. Here, ( n = 100 ), so the sum is ( S_{100} = \frac{100(100 + 1)}{2} = \frac{100 \times 101}{2} = 5050 ). Therefore, the sum of the integers from 1 to 100 is 5050.
101
It is 100*(100+1)/2 = 50500.
It is 2500.
The mean of the first 100 integers can be calculated by finding the sum of these integers and dividing by the total count. The sum of the first 100 integers (from 1 to 100) is ( \frac{100(100 + 1)}{2} = 5050 ). Dividing this by 100 gives a mean of ( \frac{5050}{100} = 50.5 ). Therefore, the mean of the first 100 integers is 50.5.
n(n+1)/2 5050
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 mean of the first 100 integers can be calculated by finding the sum of these integers and dividing by the total count. The sum of the first 100 integers (from 1 to 100) is ( \frac{100(100 + 1)}{2} = 5050 ). Dividing this by 100 gives a mean of ( \frac{5050}{100} = 50.5 ). Therefore, the mean of the first 100 integers is 50.5.
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
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.
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 integers from 100 to 1000, you can use the formula for the sum of an arithmetic series. The series has a first term (a) of 100, a last term (l) of 1000, and the number of terms (n) can be calculated as ( n = \frac{(l - a)}{d} + 1 ), where d is the common difference (1 in this case). This gives us ( n = \frac{(1000 - 100)}{1} + 1 = 901 ). The sum (S) can then be calculated using ( S = \frac{n}{2} (a + l) ), resulting in ( S = \frac{901}{2} (100 + 1000) = 450450 ). Thus, the sum of the integers from 100 to 1000 is 450450.
Write a program to find the number and sum of all integers from 100 to 300 that are divisible by 11