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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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Traxter Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

chihiroasleaf Group TitleBest ResponseYou've already chosen the best response.0
if n = 1 , n = 8
 one year ago

irudayadhason Group TitleBest ResponseYou've already chosen the best response.0
there are many numbers for N i am asking their sums
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
@Traxter gave you the sums of natural numbers
 one year ago

Traxter Group TitleBest ResponseYou've already chosen the best response.0
It might be helpful to note that the sum of n odd numbers is a perfect square: \[1+3+5+...+(2n1)=n^2\] So we want the sum of all the odd numbers betwee 1 and 999. So take (2n1)=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.
 one year ago

Traxter Group TitleBest ResponseYou've already chosen the best response.0
I think that's what you're looking for @irudayadhason.
 one year ago

irudayadhason Group TitleBest ResponseYou've already chosen the best response.0
the answer should be an integer between 0999
 one year ago

Traxter Group TitleBest ResponseYou've already chosen the best response.0
The question really doesn't make much sense, do you think you could rephrase it please?
 one year ago

irudayadhason Group TitleBest ResponseYou've already chosen the best response.0
no there are no mistakes in this question
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
i think he is looking for values of N but all the answers added up @irudayadhason
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
so we want to solve\[1+2+3+...+n=m^2\]\[n(n+1)=2m^2\]
 one year ago

irudayadhason Group TitleBest ResponseYou've already chosen the best response.0
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 0999
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
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:\[n_1+n_2+n_3+...\]
 one year ago

irudayadhason Group TitleBest ResponseYou've already chosen the best response.0
sorry 9 is a wrong answer try to use this key technique "Pell's Equation."
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
I didn't say 9 way the answer  I was using it as an example to explain the question. :)
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
the aim of this site is not "just to hand out answers". it is to teach and learn.
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
\[n(n+1)=2m^2\]\[4n(n+1)=8m^2\]\[(2n+1)^2=8m^2+1\]letting\[x=2n+1\]\[y=2m\]will gives\[x^22y^2=1\]
 one year ago

irudayadhason Group TitleBest ResponseYou've already chosen the best response.0
you mean the answer is ...
 one year ago

irudayadhason Group TitleBest ResponseYou've already chosen the best response.0
Traxter please try to solve
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
later equation solved by Pell see here http://en.wikipedia.org/wiki/Square_triangular_number
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
answer will be\[x=P_{2k}+P_{2k1}\]\[y=P_{2k}\]\(P_k\) is kth Pell number see here http://en.wikipedia.org/wiki/Pell_number
 one year ago

irudayadhason Group TitleBest ResponseYou've already chosen the best response.0
i need an answer between 0999
 one year ago
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