100
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.
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.
let the number be n-1, n, n+1 then 3n=99 and so n=33 the numbers are 32, 33, 34
100
The sum of all odd numbers 1 through 99 is 9,801.
100
4950
-99
100
99
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 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.
answer:2500
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 1+ 3 + 5 + ... + 99 = 50 x (1 + 99) ÷ 2 = 2500.
The total of all of the numbers from 1 to 99 is 4950.