2006-06-20 23:07:56 · 27 answers · asked by garnet 2 in Science & Mathematics ➔ Mathematics
The sum of all the numbers 1 through 100 is 5050.
Whenever you have to add the sum of a numbers in a consecutive interger squence starting with one; find the average of the first and last numbers, (in this case the sum is 101 and average is 50.5). Then multiply the average by the highest number (100). Therefore, 100 * 50.5 =5,050
2006-06-20 23:16:15 · answer #1 · answered by MathTeen 2 · 3⤊ 1⤋
1+2+3+4+5+6+7+8++9+10+11+12+..... +100 =you
2006-06-21 06:09:39 · answer #2 · answered by ♀guardian of angels♀ 3 · 0⤊ 0⤋
1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20....... is this a question if it is i dont get it .... =/
2006-06-21 06:11:26 · answer #3 · answered by starbucks! 2 · 0⤊ 0⤋
1+2+3+4+5+6+7+8++9+10+11+12+..... 100 = 5033
(that's really true!!!! and acurrate!!!!)
2006-06-21 06:14:32 · answer #4 · answered by laviet09 4 · 0⤊ 0⤋
1+100= 101. The same is true for 2+99; it equals 101. You could do this 50 times before you got to 51+50. Simply multiply 101 by 50 and get the correct answer, 5050.
2006-06-21 06:14:24 · answer #5 · answered by Crash&Burn 5 · 0⤊ 0⤋
n(n+1)/2
2006-06-21 06:12:55 · answer #6 · answered by alireza j 1 · 0⤊ 0⤋
13
2006-06-21 06:47:20 · answer #7 · answered by Anonymous · 0⤊ 0⤋
1 + 2 + ... + 99 + 100 = (0 + 100) + (1 + 99) + (2 + 98) + ... (49 + 51) + 50 = 5050
2006-06-21 14:01:01 · answer #8 · answered by Anonymous · 0⤊ 0⤋
1+2+3+4+5+6+7+8+9+10+11+12+..................infinity
2006-06-21 07:18:11 · answer #9 · answered by fazi 3 · 0⤊ 0⤋
For example your limit is 100
then n = 100
[n*(n+1)] / 2 = [100 * 101] / 2 = 5050
Formula works for any limit...
2006-06-21 06:34:05 · answer #10 · answered by trexx 1 · 0⤊ 0⤋