anonymous
  • anonymous
what is the sum of all even numbers between 1 and 1000
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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  • Owlfred
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anonymous
  • anonymous
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
  • anonymous
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

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anonymous
  • anonymous
@shrishti, I think you made a slight error while calculating 996/2, which should equal 498, instead you have 488.
anonymous
  • anonymous
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
  • anonymous
Another way of seeing this is sum from k =1 to k = 500 (2k) sorry about the lack of mathematics text.
anonymous
  • anonymous
\[\sum_{1}^{500}\] 2k hmmm still not good is it?
anonymous
  • anonymous
1002 * 250 = 250,500
anonymous
  • anonymous
jacobmridul check your answer again the nth term is 1000
anonymous
  • anonymous
@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.

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