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How to solve this equation?

This what I reached with my equality equation, but I could not continue: $$\frac{2n}{2} = \frac{n(n+1)}{6}$$

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Have you tried some numbers? –  Raskolnikov Apr 4 '12 at 7:48
    
Yes, do some work. Try Wolfram, it will solve it for you without any effort. –  copper.hat Apr 4 '12 at 8:12

2 Answers 2

First simplify $2n/2$ to $n$. Then multiply through by $6$ to get rid of the fractions: $6n=n(n+1)$. There are several ways to proceed from here. The neatest, perhaps, is to notice that this is certainly true when $n=0$, so that’s one solution. If $n\ne 0$, we can divide both sides by $n$ to get $6=n+1$, or $n=5$; this is the only other solution.

The straightforward, by-the-book approach is to bring everything to one side of the equation to get $n^2-5n=0$. The lefthand side factors as $n(n-5)$, so you have $n(n-5)=0$. The product of two real numbers is $0$ if and only if at least one of the two numbers is $0$, so this equation is satisfied only when $n=0$, $n-5=0$, or both $-$ i.e., when $n=0$ or $n=5$, just as we saw before. (Of course it’s not possible for $n$ and $n-5$ to be $0$ simultaneously, so the both option doesn’t apply here.)

Of course you can also use the quadratic formula to solve $n^2-5n=0$, but it would be a great waste of time and effort when the factorization of the lefthand side is so obvious.

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From you brought this n2−5n=0, (or how)? –  SolidUs Apr 4 '12 at 8:01
    
@SolidUs: You have $6n=n(n+1)$. Multiply out the righthand side: $6n=n^2+n$. Now subtract $6n$ from both sides: $0=n^2+n-6n=n^2-5n$. –  Brian M. Scott Apr 4 '12 at 8:06

2n/2 =n as the 2 gets cancelled out.

So you get n =n(n+1)/6 ie. 6n=n2 + n

now collect all the terms together on the R.H.S

n2 -5n = 0

factorizing we get :

n(n-5) = 0

Now if the product of 2 numbers is 0 then atleast one of them must be 0

so either n = 0 or n - 5 = 0

which gives you : n=0 or n=5

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