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Q: What is the sum of the first 50 positive integers?

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It can be.

They total 650. * * * * * It is 2550.

Should be 50! Every odd integer is 1 less than the corresponding even integer and there are 50 of each in 100...

sum of n natural number is n(n+1)/2 first 50 number sum is 50(50+1)/2 = 1275

810. All of the integers from -30 to30 cancel each other out by adding a negative to a positive. Then all you need to do is add the integers from 31 to 50. (31 + 32 + ....+50)

The sum of the first n positive integers (i.e. 1+2+3+...+n) is given by n(n+1)/2 so the sum of the first 25 positive integers is 1+2+3+...+25 = 25x26/2 = 325. The sum of the first 25 EVEN numbers will be double this i.e. 2+4+6+...+50 = 650.

1, 2, 3,...48, 49, 50.

Type your answer here... 25.5

The set of positive odd integers.

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.

50%

-15

The integers are 49 and 50.

24. The second is 26.

5 & 10

Sum of first n integers = n/2(n+1), in this case 25 x 51 = 1275

49

If you write down all of the integers between the two numbers, your sum is equivalent to 5,659.

5 and 10

50, 51, 52

They are: 49+50+51 = 150

no solultion!

The sum of the first 50 whole numbers is 1,225.

-26 and -24

50 + 51 + 52 = 153