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anonymous

  • 5 years ago

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

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  1. Owlfred
    • 5 years ago
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  2. anonymous
    • 5 years ago
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    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

  3. anonymous
    • 5 years ago
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    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

  4. anonymous
    • 5 years ago
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    @shrishti, I think you made a slight error while calculating 996/2, which should equal 498, instead you have 488.

  5. anonymous
    • 5 years ago
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    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

  6. anonymous
    • 5 years ago
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    Another way of seeing this is sum from k =1 to k = 500 (2k) sorry about the lack of mathematics text.

  7. anonymous
    • 5 years ago
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    \[\sum_{1}^{500}\] 2k hmmm still not good is it?

  8. anonymous
    • 5 years ago
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    1002 * 250 = 250,500

  9. anonymous
    • 5 years ago
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    jacobmridul check your answer again the nth term is 1000

  10. anonymous
    • 5 years ago
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    @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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