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Given $(u+1) \le n^2$, I am trying to understand under what circumstances:

$\left[(\frac{u+1}{2n})^{2n} - (\frac{u}{2n})^{2n}\right]$ < $\left[(u+1)^n-u^n\right]$

It seems like it should be easy to prove but I'm not sure where to begin.

At this point, I am thinking that the approach requires using the Binomial Theorem? Is that right? Or is it enough to figure out the first derivative?

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\begin{align} \left[ \left( \frac{u+1}{2n}\right)^{2n}-\left(\frac{u}{2n} \right)^{2n}\right] &= \frac{1}{(2n)^{2n}} \left[\left((u+1)^n \right)^2 - (u^n)^2\right] \\ &=\frac{1}{(2n)^{2n}} \left[\left((u+1)^n -u^n\right) ((u+1)^n+u^n)\right] \\ &=((u+1)^n-u^n)\left[\left( \frac{u+1}{4n^2}\right)^n+\left(\frac{u}{4n^2}\right)^n\right]\\ \end{align}

A sufficient condition is when $u,n \ge 0$.

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