The total T of adding numbers from 1 to n is (n x (n+1))/2; consider if n=10:
adding 1+ 2+ 3+...+ 8+ 9+ 10= T and
adding 10+ 9+ 8+...+ 3+ 2+ 1= T (the same answer)
Now add the the two lines together--->
11+11+11+..+11+11+11= 2T, so there are 10 11s
10 x 11 =2T, divide both sides of the equal sign by 2
(10 x 11)/2 =T
So, where n=10, T=55
Consider that 1+99=100 and 2+98=100 you can continue all the way to 49+51=100. This means you have (49x100)+50 (as fifty was left at the end) this gives a final answer of 4950
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.
The sum of all the odd numbers from 1 through 100 is 10,000
The sum of all the first 100 even numbers is 10,100.
This is attributed to an early school lesson when the teacher thought he would keep the class busy whilst he popped out for something. He set the test of adding all the whole numbers from 1-100. By the time he reached the door, Gauss had the answer. Gauss imagined the problem as 1 + 2 + 3 +........+98 + 99 + 100, but then he wrote the numbers underneath but in reverse order. 100 +99 + 98..........+3 + 2 + 1 So each 100 pairs of vertical numbers added up to 101 so the total was 10100 but this is twice the true answer as each number is included twice. The total is therefore 5050. This lead to the general formula that the sum of consecutive numbers from 1 to n is n(n+ 1) ÷ 2.
The total of all of the numbers from 1 to 99 is 4950.
A scale is the total of all your numbers added up.
Add up all the numbers. Divide that total by the number of numbers you added.
All the numbers from 100 to 999 have three digits, as do all the numbers from -100 to -999, so that's 1800 in total.
- To get the arithmetic mean you add all the numbers up. Then, count the number of numbers you added, and divide the sum by the total number of numbers.
The odd numbers 1-99 will be a higher total than the even numbers 2-98
5050, ie (1 + 100) x (100/2)
The sum of the first 100 positive even numbers is 10,100.
Imagine all of the 100 numbers from 1 to 100 written horizontally towards the right. Then imagine another100 numbers written below the first row, but starting with 100 at the left and finishing with the number 1 on the right hand end. Now you can see what looks like 100 additions of two numbers. If you worked out those additions they would all total exactly 101, which means that altogether we have 100 pairs of numbers each totalling 101, and therefore the grand total of all 200 numbers is 101 x 100 = 10100; but 10100 is the total of two sets of numbers from 1 to 100; we want the total of one set of numbers; so we divide the 10100 by 2 which is 5050; the beauty of this method is that you don't need complicated calculations; all you need to be able to do is to multiply by 100 and to divide by 2; a very neat solution to the problem!
The average is all the numbers added together and divided by how many numbers there are. In this case: 78+100=178. 178/2=89. Therefore the mean/average of the two numbers is 89.
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.
You can either just add them up or use quick logic to add them in pairs: All the numbers from 1 to 49 match up with 51 through 99 to = 49 pairs of 100 = 4900, than add the two you skipped 4900 + 50 + 100 = 5050 total