n=100
{[(n/4)-1]*n}+(n/2)
={[(100/4)-1]*100}+(100/2)
={[25-1]*100}+50
={24*100}+50
=2400+50
=2450
2, 4, 6, 8, 10, 12 and just keep adding 2 until you get to 100.
The even numbers up to 100 are: 2 468101214161820222426283032343638404244464850525456586062646668707274767880828486889092949698100.
even numbers 1-100
To find the even numbers between 100 and 400, we note that the range includes numbers from 102 to 398. The first even number is 102, and the last is 398. The even numbers form an arithmetic sequence with a common difference of 2. To find the count, we can use the formula for the number of terms in an arithmetic sequence: ( n = \frac{(last - first)}{difference} + 1 ), which gives ( n = \frac{(398 - 102)}{2} + 1 = 149 ). Thus, there are 149 even numbers between 100 and 400.
10100, not sure how you do it other than the incredibly long way of adding every single number but I'm pretty sure thats the answer
2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50,52,54,56,58,60,62,64,66,68,70,72,74,76,78,80,82.84,86,88,90,92,94,96,98,and 100Start with 2 and keep adding 2. JHC!
2, 4, 6, 8, 10, 12 and just keep adding 2 until you get to 100.
Adding up all the even numbers from the first (2) to the fiftieth (100): [(100 + 2) / 2] x 50 = 2550 Can you see the logic of this method?
The first 100 even numbers are: 2 468101214161820222426283032343638404244464850525456586062646668707274767880828486889092949698100102104106108110112114116118120122124126128130132134136138140142144146148150152154156158160162164166168170172174176178180182184186188190192194196198200.
The first 100 whole, even numbers are: 2468101214161820222426283032343638404244464850525456586062646668707274767880828486889092949698100102104106108110112114116118120122124126128130132134136138140142144146148150152154156158160162164166168170172174176178180182184186188190192194196198200.
The even numbers up to 100 are: 2 468101214161820222426283032343638404244464850525456586062646668707274767880828486889092949698100.
even numbers 1-100
The sum of the first 100 positive even numbers can be calculated using the formula for the sum of an arithmetic series: n*(first term + last term)/2. In this case, the first term is 2, the last term is 200, and n is 100. Therefore, the sum is 10,100.
Adding: 100+44 Multiplying: 12*12
The sum of the first 100 positive even numbers is 10,100.
In Java:System.out.println("Even numbers")for (int i = 2; i
To find the even numbers between 100 and 400, we note that the range includes numbers from 102 to 398. The first even number is 102, and the last is 398. The even numbers form an arithmetic sequence with a common difference of 2. To find the count, we can use the formula for the number of terms in an arithmetic sequence: ( n = \frac{(last - first)}{difference} + 1 ), which gives ( n = \frac{(398 - 102)}{2} + 1 = 149 ). Thus, there are 149 even numbers between 100 and 400.