Prove that $\sum_{k=n^2+1}^{n^2+2n+1}\sqrt{k}\le 2n^2+2n+\frac{7}{6}$ prove that:
$$\displaystyle\sum_{k=n^2+1}^{n^2+2n+1}\sqrt{k}\le 2n^2+2n+\dfrac{7}{6},n\ge 1$$
I find  this inequality is very strong. Thank you!
such:when $n=100$,we use the mathmatic 
$$\displaystyle\sum_{k=10001}^{10201}\approx 20201.16666254125412794937\cdots $$
and $2n^2+2n+\dfrac{7}{6}|_{n=100}=20201.166666666666\cdots$

Note that your bound gets much tighter as $n$ increases, suggesting that some kind of limiting argument is involved.
With that insight, we have that
$$ \begin{align} 
\sum_{k = n^2 + 1}^{k=n^2+2n+1} \sqrt{k} & = \sum_{k=n^2} ^{k=n^2 + 2n} \sqrt{k} + 1 \\ 
& = 1 + \sum_{k=n^2} ^{k=n^2+2n} \int_k^{k+1} \sqrt{k} \, dx \\ 
& \leq 1 + \sum_{k=n^2} ^ {k=n^2 + 2n} \int_{k}^{k+1} \sqrt{x} \, dx \\ 
& = 1 + \int_{n^2} ^{(n+1)^2} \sqrt{x} \, dx\\
& = 1 + [ \frac {2}{3} k^{\frac {3}{2}} ]_{n^2} ^{(n+1)^2} \\
& = 1 + [\frac {2}{3} (3n^2 + 3n+1) ] \\
& = 2n^2 + 2n + \frac {5}{3}
\end{align}
$$
Now, let's consider how much we over approximated by. Looking at the inequality, we over approximated by $\int_{k}^{k+1} \sqrt{x} - \sqrt{k}$. Since the square root function is concave, we have the extra triangle on top, which tells us that we over approximated by at least $\frac {1}{2} (\sqrt{k+1}-\sqrt{k})$. Summing this up, we get 
$$\sum_{k=n^2}^{n^2+2n} \frac {1}{2} ( \sqrt{k+1}- \sqrt{k} ) = \frac {1}{2} ( k+1 - k) = \frac {1}{2}. $$
Hence, (back)accounting for this over-approximation, we know that 
$$ \sum_{k = n^2 + 1}^{k=n^2+2n+1} \sqrt{k} \leq  2n^2 + 2n + \frac {5}{3} - \frac {1}{2} = 2n^2 + 2n + \frac {7}{6}. $$
