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irudayadhason
What is the sum of all values of N, 1≤N≤999, such that 1+2+3+…+N is a perfect square?
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?
there are many numbers for N i am asking their sums
@Traxter gave you the sums of natural numbers
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.
I think that's what you're looking for @irudayadhason.
the answer should be an integer between 0-999
The question really doesn't make much sense, do you think you could rephrase it please?
no there are no mistakes in this question
i think he is looking for values of N but all the answers added up @irudayadhason
so we want to solve\[1+2+3+...+n=m^2\]\[n(n+1)=2m^2\]
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
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:\[n_1+n_2+n_3+...\]
sorry 9 is a wrong answer try to use this key technique "Pell's Equation."
I didn't say 9 way the answer - I was using it as an example to explain the question. :)
the aim of this site is not "just to hand out answers". it is to teach and learn.
\[n(n+1)=2m^2\]\[4n(n+1)=8m^2\]\[(2n+1)^2=8m^2+1\]letting\[x=2n+1\]\[y=2m\]will gives\[x^2-2y^2=1\]
you mean the answer is ...
Traxter please try to solve
later equation solved by Pell see here http://en.wikipedia.org/wiki/Square_triangular_number
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
i need an answer between 0-999