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Q: What is the sum of the numbers 1 to 99?

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The sum of all odd numbers 1 through 99 is 9,801.

4950

100

-99

99

100

The sum of 1+ 3 + 5 + ... + 99 = 50 x (1 + 99) ÷ 2 = 2500.

answer:2500

The total of all of the numbers from 1 to 99 is 4950.

The sum of the integers 1 to 99 is 4950. An easy way to figure this out is using the equation N*(N+1)/2 where N is the largest number in the set.

let the number be n-1, n, n+1 then 3n=99 and so n=33 the numbers are 32, 33, 34

9801 * The sum of the first two odd numbers (1+3) is 4, or 22 * The sum of the first three odd numbers (1+3+5) is 9, or 32 * The sum of the first four odd numbers (1+3+5+7) is 16, or 42 * ...and so on So the sum of the first 99 odd numbers, using the pattern above, would be 992 or 9801.

The sum of the even numbers is (26 + 28 + ... + 100); The sum of the odd numbers is (25 + 27 + ... + 99) Their difference is: (26 + 28 + ... + 100) - (25 + 27 + ... + 99) = (26 - 25) + (28 - 27) + ... + (100 - 99) = 1 + 1 + ... + 1 There are (100 - 26) ÷ 2 + 1 = 38 terms above which are all 1; their sum is 38 x 1 = 38. So the difference of the sum of all even numbers and all odd numbers 25-100 is 38.

99^2 == 9,801

the series of odd numbers from 1 to 99 :1 3 5 7 9.....99 SUM OF THE SERIES: It is a geometric progression with a=1 and l=99 and common difference (d)=2. let 99 be the nth term of the sequence so, 99=1+(n-1)d 99=1+(n-1)2 solving this we get n =50. SUM=(n/2)(a+l) =(50/2)(1+99) =(25)(100) =2500.

2500

100/3-1, 100/3 and 100/3+1 that is, 99, 100 and 101.

n/2(n + 1) = 49½ x 100 = 4950

Since 1+99=100. And 2+98=100 and 3+97=100 etc to 49+51=100 it's 49 of those plus 50 or 4,950. The sum of the first 99 whole numbers is 4,950

Do you mean:"What three consecutive numbers add up to 99?" If you do, then the numbers are 32, 33, 34.

101

First we find the sum of the integers from 2 to 100 including 2 and 100, then we divide by 99 since there are 99 numbers. There are 100 numbers between 1 and 100 and we are excluding only the number 1. If we want to exclude the number 2, and count only the numbers 3,4,5....100, this can be done with the same procedure and a slight modification. So the sum of the numbers 1 to 100 can be found by writing the numbers 1, 2,3,...100 Now write them backwards, starting at 100,99,98....1 Each column has a sum of 101 and there are 100 columns. So the total is 100x101, but we wrote the list twice so we must divide by 2 The sum is 100x101/2=50x101=5050 Now remember we want only 2 to 100 so the sum we seek is 5049. Since there are 99 number, the mean is 5049/99=51 In general the sum of the first n positive integers is n(n+1)/2 This can be proved the way we did or by induction.

1. The numbers 1 through 99 are 99 different numbers; each one occurs once.

The numbers are 97, and 99.

The sum of the numbers from 1 through 100 is 5,050.