Given a number $x$, how do we find a value $n$ such that $$ 1+2+3+...+n =x $$ This is how far I've gotten: $$ x=\frac{n(n+1)}{2} $$ $$ 2x = n(n+1) $$ What do I do from here???
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asked Mar 23, 2018 at 11:43
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16$\begingroup$ Hint: Quadratic formula $\endgroup$– luluMar 23, 2018 at 11:43
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1$\begingroup$ @lulu Oh....... $\endgroup$– Aravind Suresh ThakidayilMar 23, 2018 at 11:45
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2 Answers
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Then we will have $n^2+n-2x=0$
You can find solution for $n$ if and only if $x$ is a natural number and $\Delta=1^2-4\times 1 \times (-2x)=8x+1\ge0$, which is true for all nautral numbers $x$.
Then we will have $n=\frac{-1+\sqrt{8x+1}}{2}$ or $n=\frac{-1-\sqrt{8x+1}}{2}$, but we will only take the positive integer solution, eliminate any non-integer solution.
Note that $8x+1$ must be a perfect square number so that $n$ can be a positive integer.
answered Mar 23, 2018 at 11:56
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1$\begingroup$ Note that $8x+1$, if a square, is always going to be odd, so specifying that isn't necessary. $\endgroup$– user304329Mar 23, 2018 at 11:59
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$\begingroup$ So $n=\max(\frac{-1+\sqrt{8x+1}}{2}, \frac{-1-\sqrt{8x+1}}{2})$? $\endgroup$ Mar 23, 2018 at 12:00
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$\begingroup$ Note that $8x+1$ will always be odd, so the only condition is that it is a square. $\endgroup$ Mar 23, 2018 at 12:00
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$\begingroup$ @user5011: Well, yes, but it's simpler just to say $n=\frac{-1+\sqrt{8x+1}}2.$ $\endgroup$ Mar 23, 2018 at 12:01
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$\begingroup$ Okay, got it now. Thanks a lot!$n = \frac{-1+\sqrt{8x+1}}{2}$. $\endgroup$ Mar 23, 2018 at 12:02
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You need to solve $$ n^2+n-2x =0$$
As you know this quadratic equation may or may not have integer solutions.
For example $ x=55$ results in $n=10$ but $x=50$ does not have a round solution.
answered Mar 23, 2018 at 12:12