what is the sum of all even numbers between 1 and 1000

Question

owlfred · Answer

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anonymous · Answer

sum of all even number between 1 and 1000 is 2+4+6.......................................998
a=2
d=2
an = a+ (n-1)d
998= 2+n-1)2
998-2=(n-1)2
996=(n-1)2
488=n-1
n=489
s489= 489/2[2+489]
           =489/2[491]=120049

anonymous · Answer

That's the sum of all terms of the  arithmetic progression, with 
a=2 and d=2. last term is 998
a=2
d=2
nth term = 998

998=a+(n-1)d
998=2+(n-1)2
998=2n
n=499
 sum=n/2(2a+(n-1)d)

 therefore sum of 499 terms =  249500

anonymous · Answer

@shrishti, I think you made a slight error while calculating 996/2, which should equal 498, instead you have 488.

anonymous · Answer

There are 500 even numbers
first is 2 last is 1000
second is 4 last but one is 998
You will see (just like Gauss did ) that they both sum to 1002
So as there are 250 of these 'sums' ...
250 x 1002 = 250 500
This is a correct answer

anonymous · Answer

Another way of seeing this is  sum from k =1 to k = 500  (2k) sorry about the lack of mathematics text.

anonymous · Answer

$$\sum_{1}^{500}$$ 2k 

hmmm still not good is it?

anonymous · Answer

1002 * 250 = 250,500

anonymous · Answer

jacobmridul check your answer again the nth term is 1000

anonymous · Answer

@gianfranco, I'm afraid the question is sum of all even numbers "BETWEEN"  1 and 1000, in my humble understanding it excludes 1000. so the number of terms is 499 not 500.