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$