## irudayadhason 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

1. Traxter

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. chihiroasleaf

if n = 1 , n = 8

there are many numbers for N i am asking their sums

@Traxter gave you the sums of natural numbers

5. Traxter

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.

6. Traxter

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

the answer should be an integer between 0-999

8. Traxter

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

no there are no mistakes in this question

11. mukushla

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

13. asnaseer

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."

15. asnaseer

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

16. asnaseer

*was

17. asnaseer

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

18. mukushla

$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 ...

21. mukushla

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

22. mukushla

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