100
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.
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.
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 all odd numbers 1 through 99 is 9,801.
100
answer:2500
Well, darling, the sum of the first 99 odd numbers is 9801. You see, you take the last odd number, which is 197, multiply it by the total number of odd numbers (which is 99), and divide by 2. Voila! Math magic at its finest.
100
99
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.
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.
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.
let the number be n-1, n, n+1 then 3n=99 and so n=33 the numbers are 32, 33, 34