What is the sum of 1+2+3+4....+999+1000?

Question

fastest correct answer wins!!!

I believe it's 5,005 is that correct?

Answer 1

500500

Answer 2

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As u can see n th term (Tn) = 1 / [(n)(n+1)] Find partial fractions as follows . . 1. . . . . . . A. . . . . . . B - - - - - - = - - - - - + - - - - - - - - n(n + 1). . . . n. . . . . .(n + 1) 1 = A(n+1) + B(n) A = 1 B = - 1 so; . . 1. . . . . . . 1. . . . . . . 1 - - - - - - = - - - - - _ - - - - - - - - n(n + 1). . . . n. . . . . .(n + 1) let; 1/n = Vn so 1/(n+1) = V(n+1) T1 = V1 - V2. . . . . . . . .| T2 = V2 - V3. . . . . . . . .| T3 = V3 - V4. . . . . . . . .| T4 = V4 - V5. . . . . . . . .| : : : : : : :: : :. . . . . . . . . | T(n-1) = V(n-1) + Vn. . . .| Tn = Vn - V(n-1) . . . . . .| . . . . . . . . . . . . . . . . . ▼ = = = = = = = = = = = = ∑ (n=1 to n) 1/[(n)(n+1)] = V1 - V(n+1) Sn = 1/1 - 1/(n+1) So if Tr = 1/[(r)(r+1)] n. . . . . . 1. . . . . 1 ∑. . Tr = - - - _ - - - - - r=1. . . . .1. . . . (n+1) in short Sum to the n th term = 1 - 1/(n+1) Sum to the n th term = n/(n+1) so 1/(999)(1000) is the 999th term so sum to 999 th term = (999)/(999+1) = 999/1000 = 0.999 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = That was the hard way. U can do this just by lookig at some terms S1 = 1/2 S2 = 1/2 + 1/6 = 2/3 S3 = 1/2 + 1/6 + 1/12 = 3/4 S5 = 1/2 + 1/6 + 1/12 + 1/20 = 4/5 So for s999 = 999/1000 = 0.999 That'd be the way to do it for a multiple choice question.

Answer 3

Sum Of 1 To 1000

Answer 4

A teacher, wanting to keep her students busy for a while gave this same question to her class. It so happenened that one of her students was Gauss, who became a famous mathemetician

To the teachers astonisment young Gauss answered the question in just a few minutes. His reasoning was:
1000 +1 =1001
999 +2 =1001
998 +3 =1001
etc

Hes saw that bthere would be 500 pairs of numbers that added to 1001 so he multiplied 500* 1001 to get the answer 500,500.

He had essentially created the formula (n/2)(n+1) where n is nthe total number of terms. Not bad for a yuung kid, eh?

Answer 5

sum of terms = number of terms/2 * (first + last term)

let 1+2+3+4....+999+1000 = S
s= 1000/2(1+1000)
s= 500*1001
= 500500

Answer 6

sum of terms = number of terms/2 x (first + last term)

1+2+3+4....+999+1000 = 1000/2(1+1000)
500*1001= 500500

Answer 7

who cares about winning a measly 10 points, especially when those points have no market value whatsoever? Get real! ;-)

anyway, you dont' seem to know there's a formula for this, the sum of numbers to "n", it is n*(n+1)/2

in this case, that's 1000*1001/2, which is 500'500

Answer 8

n(n+1)/2=1000*1001/2=500500

Answer 9

use the equatio sum=(n/2)(a+l)

n=nomber of nobbers
a=first nomber
l=larst nomber
sum=(1000/2)(1000+1)
=500*1001
=500500

Answer 10

the correct answer is 500,500