2007-12-29 12:33:17 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
1 + 2 + ... + 50 +
100 + 99 + ... + 51
add them vertically
50 lots of 101 = 5050
Or just use the formula n(n+1)/2.
Or do it is an arithmetic series.
2007-12-29 12:38:06 · answer #1 · answered by Raichu 6 · 1⤊ 0⤋
1 + 2 + 3 + 4 + 5 + ... ... + 98 + 99 + 100
= (1 + 100) x 100 / 2
= 101 x 100 / 2
= 10100 / 2
= 5050
(first term + last term) x total terms / 2
2007-12-29 20:18:51 · answer #2 · answered by An ESL Learner 7 · 0⤊ 0⤋
n(n+1)/2, where n = 100 in this case
2007-12-29 17:00:30 · answer #3 · answered by fraukka 3 · 0⤊ 0⤋
you will notice that as you increase the first term, and decrease the second, you will get 101 (1 + 100, 2 + 99, etc) 50 times. So 5,050 is the answer.
2007-12-29 12:42:17 · answer #4 · answered by Charles M 6 · 0⤊ 0⤋
FYI, the sum of the first n positive integers is given by the formula n*(n+1)/2. (So n doesn't have to be 100 for n*(n+1)/2 to work.)
2007-12-29 13:01:36 · answer #5 · answered by AxiomOfChoice 2 · 0⤊ 1⤋
if you add 100 (last numbeR) and 1( first number) you get a total of 101 and all other numbers until you get to 50 and 51 (50th numberS) this means that there are 50 sums of 101 so if you multiply them you will get 5050
2007-12-29 12:38:28 · answer #6 · answered by ViewtifulJoe 4 · 0⤊ 0⤋
Ask Karl Friedrich Gauss
2007-12-29 13:51:12 · answer #7 · answered by Mary E 3 · 0⤊ 0⤋
use a calculator its 5050
2007-12-29 12:42:26 · answer #8 · answered by Giggles 2 · 0⤊ 1⤋