Question

1-2-3-4-5-...-999=?

Answer 1

 The problem is to find general formula for this geometric progression. 

Example : 1 + 2 + 3 + 4 + ... +8+9 + 10 (reorganized the serie's 
terms) 
= (10+1) + (2+9) + (3+8) + (4+7) + (5+ 6) 
= 11 + 11 + 11 + 11 + 11 
= 11 x 5 
= 55 

So, Sn = 1 + 2 + 3 + ... (n-3) + (n-2) + (n-1) + n 
Sn = (n+1) + (n-1 +2) + (n-2 + 3) + etc 
Sn = (n + 1) + (n+1) + (n +1) + ...+ (n+1) (n/2) times 
Sn = (n/2)(n+1) 
Sn = (n^2 + n)/2 is the general formula(easier way to find the 
answer) 

Example: S(10) = ((10^2) + 10)/2 
= (100 + 10)/2 
= 55 

So, S(1000) = ((1000)^2 + 1000)/2 
= (1000000 + 1000)/2 
= 1001000/2 
= 500500 => answer 

dựa vào nhé

Answer 2

kho nen khong bit lam