A PuiChing Question (to someone)

Problem. Denote $[x]$ the greastest integer not exceeding $x$, calculate $\displaystyle\sum\limits_{k=1}^{100}[\sqrt{k(k+4)+20}]$.

Solution.
First note that $\sqrt{k(k+4)+20}=(k+2)\sqrt{1+\big(4/(k+2)\big)^2}$, it conveys us some messages. As $k$ increases, we have $\sqrt{k(k+4)+20}= k+2 + f(k)$, where $f(k)\to 0$ as $k\to\infty$. But only the integral part deserves our concern, so if $k+2 <\sqrt{k(k+4)+20}<k+3$, then everything goes smooth and we have $[\sqrt{k(k+4)+20}]=k+2$. But when does this inequality hold? Just solve the inequality we have $k>5.5$, so the inequality is true of $k\ge 6$ and thus the value should be $5+5+6+7+8+8+9+10+\cdots +102 = 5256$.     $\blacksquare$