99^2 == 9,801
100
No. As both negative and positive numbers can be odd, there is no first odd number, and therefore no 100th odd number. The 100th odd positive number is 199.
Let n = smallest of the odd numbers, then let n+2 = the larger of the two numbers (Remember, 1 is not a prime number.) n+ n+2 = {(2)(7)}2 2n +2 = 142 2n = 196 -2 2n = 194 n = 97 n + 2 = 99
Infinity
50 of them.
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 all odd numbers 1 through 99 is 9,801.
100
2500
100
99
answer:2500
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.
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.
The sum of 1+ 3 + 5 + ... + 99 = 50 x (1 + 99) ÷ 2 = 2500.
let the number be n-1, n, n+1 then 3n=99 and so n=33 the numbers are 32, 33, 34
The sum of 50 consecutive odd integers is equal to: (First Term + Last Term) * (The number of Terms)/2 or in this case, more specifically: (2 * First Term + 98) * 25 If you meant 1+3+5+7+...+99, then the sum is 2500.