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Q: What is the sum of the first 99 odd numbers?

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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

99

100

answer:2500

99 is an odd number. Even numbers are evenly divisible by two, while odd numbers are not.

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

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.

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 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.

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.

printf ("%d\n", (1+99)/2*50).

The composite odd numbers from 9 to 99.

99

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

Their sum is 99.

The sum of all the integers between 1 and 99 inclusive is 4950.

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

The numbers are 31, 33 and 35.

4950

Infinity

Since we are talking for the sum of digits, then we are talking about 2-digit numbers between 1 and 100, which are 90 numbers (10 to 99). From these numbers only half of them satisfy the above condition, that is 45 numbers. So 50% of 2-digit numbers between 1 and 100 have an odd sum of their digits.

n/2(n + 1) = 49½ x 100 = 4950