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irudayadhason Group Title

What is the sum of all values of N, 1≤N≤999, such that 1+2+3+…+N is a perfect square?

  • 2 years ago
  • 2 years ago

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  1. Traxter Group Title
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    The sum n n numbers from 1 to n is given by: \[\frac{n(n+1)}{2}\] Are you asking for a value of N such that the sum is a perfect square, or for the sum of numbers from 1 to 999?

    • 2 years ago
  2. chihiroasleaf Group Title
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    if n = 1 , n = 8

    • 2 years ago
  3. irudayadhason Group Title
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    there are many numbers for N i am asking their sums

    • 2 years ago
  4. Jonask Group Title
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    @Traxter gave you the sums of natural numbers

    • 2 years ago
  5. Traxter Group Title
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    It might be helpful to note that the sum of n odd numbers is a perfect square: \[1+3+5+...+(2n-1)=n^2\] So we want the sum of all the odd numbers betwee 1 and 999. So take (2n-1)=999, so that 2n=1000 and n=500. So The sum of all the odd numbers between 1 and 999 is \(500^2\)=250,000.

    • 2 years ago
  6. Traxter Group Title
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    I think that's what you're looking for @irudayadhason.

    • 2 years ago
  7. irudayadhason Group Title
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    the answer should be an integer between 0-999

    • 2 years ago
  8. Traxter Group Title
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    The question really doesn't make much sense, do you think you could rephrase it please?

    • 2 years ago
  9. irudayadhason Group Title
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    no there are no mistakes in this question

    • 2 years ago
  10. Jonask Group Title
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    i think he is looking for values of N but all the answers added up @irudayadhason

    • 2 years ago
  11. mukushla Group Title
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    so we want to solve\[1+2+3+...+n=m^2\]\[n(n+1)=2m^2\]

    • 2 years ago
  12. irudayadhason Group Title
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    no i am not looking for the values of N i am just asking the sum of all values of N but it should be between 0-999

    • 2 years ago
  13. asnaseer Group Title
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    Maybe this will help explain it to others better: lets say the only solutions to this were n=1 and n=8, then your answer would be 1 + 8 = 9. so, from all the valid solutions to this (\(n_1,n_2,n_3,...\)), the answer we want is:\[n_1+n_2+n_3+...\]

    • 2 years ago
  14. irudayadhason Group Title
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    sorry 9 is a wrong answer try to use this key technique "Pell's Equation."

    • 2 years ago
  15. asnaseer Group Title
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    I didn't say 9 way the answer - I was using it as an example to explain the question. :)

    • 2 years ago
  16. asnaseer Group Title
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    *was

    • 2 years ago
  17. asnaseer Group Title
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    the aim of this site is not "just to hand out answers". it is to teach and learn.

    • 2 years ago
  18. mukushla Group Title
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    \[n(n+1)=2m^2\]\[4n(n+1)=8m^2\]\[(2n+1)^2=8m^2+1\]letting\[x=2n+1\]\[y=2m\]will gives\[x^2-2y^2=1\]

    • 2 years ago
  19. irudayadhason Group Title
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    you mean the answer is ...

    • 2 years ago
  20. irudayadhason Group Title
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    Traxter please try to solve

    • 2 years ago
  21. mukushla Group Title
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    later equation solved by Pell see here http://en.wikipedia.org/wiki/Square_triangular_number

    • 2 years ago
  22. mukushla Group Title
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    answer will be\[x=P_{2k}+P_{2k-1}\]\[y=P_{2k}\]\(P_k\) is kth Pell number see here http://en.wikipedia.org/wiki/Pell_number

    • 2 years ago
  23. irudayadhason Group Title
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    i need an answer between 0-999

    • 2 years ago
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