Sums $\sum_{n=1}^{N}\sqrt{4n+1}$ I need to find sum of the first N terms of the sequence whose nth term is as follow :
T(n)= $\sqrt{4*n+1}$
So the sequence is : $\sqrt{5}$,$\sqrt{9}$,$\sqrt{13}$,$\sqrt{17}$,$\sqrt{21}$......
Please help how to find it for a given N.

There is an answer which is not simple at all since $$\sum_{i=1}^n \sqrt{4i+1}=2 \left(\zeta \left(-\frac{1}{2},\frac{5}{4}\right)-\zeta
   \left(-\frac{1}{2},n+\frac{5}{4}\right)\right)$$ in which appears Hurwitz $\zeta$ function.
However, you could notice that $$2\sqrt{i}<\sqrt{4i+1}<2\sqrt{i+1}$$ and so  $$2H_{n}^{\left(-\frac{1}{2}\right)}<\sum_{i=1}^n \sqrt{4i+1}<2H_{n+1}^{\left(-\frac{1}{2}\right)}$$ where appear the generalized harmonic numbers. 
For large values of $m$, an approximation is $$H_{m}^{\left(-\frac{1}{2}\right)}=\frac{2 m^{3/2}}{3}+\frac{\sqrt{m}}{2}+\zeta
   \left(-\frac{1}{2}\right)+O\left(\left(\frac{1}{m}\right)^{3/2}\right)$$
For $n=10$, the exact summation leads to $46.1629$ while the expression on the left is $44.9102$ and the one on the right is $51.5447$
Added later
The approximation can be improved writing $$\sqrt{4i+1}=2 \sqrt{i}+\frac{\sqrt{\frac{1}{i}}}{4}-\frac{1}{64}
   \left(\frac{1}{i}\right)^{3/2}+O\left(\left(\frac{1}{i}\right)^{5/2}\right)$$ and expanding again the harmonic numbers. As a result comes the approximation  $$\sum_{i=1}^n \sqrt{4i+1}\approx \frac{4 n^{3/2}}{3}+\frac{3 \sqrt{n}}{2}+\left(2 \zeta \left(-\frac{1}{2}\right)+\frac{\zeta
   \left(\frac{1}{2}\right)}{4}-\frac{\zeta \left(\frac{3}{2}\right)}{64}\right)+O\left(\left(\frac{1}{n}\right)\right)$$ which does not seem to be too bad $(46.0854$ for $n=10)$.
For $n=1000$, the value of the sum is $42210.32452$ while the approximation gives $42210.31462$.
