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I am currently trying to find a pattern in $x^2+x.$

I am finding that there is a set of multiplicands that show up whenever I lay down numbers $1$ to $10$ in this formula. Make note, I am calculating how much you multiply for example $2^2-2$ to get to $3^2-3$.

These multiplicands are $3$ for $2$, $2$ for $3$, $1 \frac{2}{3}$ for $4$, $1.5$ for $5$, $1.4$ for $6$, $1\frac{1}{3}$ for $7$, and so on. I tried to see if there was a logarithmic or exponential connection between these, or something that could at least predict an outcome, and I failed. I am wondering if there is a certain more simple way to simplify or describe this behavior, or this is a stupid questions like I think it is. Thank you!

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    it is just $$x(x+1)$$ – 2017-02-17
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    Hint: $\;\require{cancel}\cfrac{(x+1)^2+x+1}{x^2+x}=\cfrac{\cancel{(x+1)}(x+2)}{x\cancel{(x+1)}}=\cfrac{x+2}{x}=1+\cfrac{2}{x}\,$ – 2017-02-17
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    you have forgotten $$x\ne 0$$ and $$x\ne -1$$ – 2017-02-17

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It's not a stupid question but it isn't a hard one either.

You can see the pattern much more clearly if you use fractions rather than "mixed numbers": $$ \frac{3}{1}, \frac{4}{2}, \frac{5}{3}, \frac{6}{4}, \ldots $$

A little algebra will prove the pattern continues:

You want to find the unknown number $?$ such that $$ (n^2+n) \times ? = (n+1)^2 + (n+1) $$ so $$ ? = \frac{(n+1)^2 + (n+1)}{n^2+n} . $$ Can you finish? Factor the numerator and denominator and see what cancels.

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    n^2+n+2/n^2+n = 2. – 2017-03-06