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

Question

anonymous · Answer

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?

chihiroasleaf · Answer

if n = 1 , n = 8

anonymous · Answer

there are many numbers for N i am asking their sums

anonymous · Answer

@Traxter gave you the sums of natural numbers

anonymous · Answer

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.

anonymous · Answer

I think that's what you're looking for @irudayadhason.

anonymous · Answer

the answer should be an integer between 0-999

anonymous · Answer

The question really doesn't make much sense, do you think you could rephrase it please?

anonymous · Answer

no there are no mistakes in this question

anonymous · Answer

i think he is looking for  values of N but all the answers added up  @irudayadhason

anonymous · Answer

so we want to solve$$1+2+3+...+n=m^2$$$$n(n+1)=2m^2$$

anonymous · Answer

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

asnaseer · Answer

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:$$n_1+n_2+n_3+...$$

anonymous · Answer

sorry 9 is a wrong answer try to use this key technique "Pell's Equation."

asnaseer · Answer

I didn't say 9 way the answer - I was using it as an example to explain the question. :)

asnaseer · Answer

*was

asnaseer · Answer

the aim of this site is not "just to hand out answers". it is to teach and learn.

anonymous · Answer

$$n(n+1)=2m^2$$$$4n(n+1)=8m^2$$$$(2n+1)^2=8m^2+1$$letting$$x=2n+1$$$$y=2m$$will gives$$x^2-2y^2=1$$

anonymous · Answer

you mean the answer is ...

anonymous · Answer

Traxter please try to solve

anonymous · Answer

later equation solved by Pell see here http://en.wikipedia.org/wiki/Square_triangular_number

anonymous · Answer

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

anonymous · Answer

i need an answer between 0-999