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

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

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

- schrodinger

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

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

if n = 1 , n = 8

- anonymous

there are many numbers for N i am asking their sums

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

@Traxter gave you the sums of natural numbers

- anonymous

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

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

- anonymous

the answer should be an integer between 0-999

- anonymous

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

- anonymous

no there are no mistakes in this question

- anonymous

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

- anonymous

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

- anonymous

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

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

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

- asnaseer

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

- asnaseer

*was

- asnaseer

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

- anonymous

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

- anonymous

you mean the answer is ...

- anonymous

Traxter please try to solve

- anonymous

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

- anonymous

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

i need an answer between 0-999

Looking for something else?

Not the answer you are looking for? Search for more explanations.