問
簡單
是也。
如果變成中四 M2 的 MI 習題：
利用數學歸納法證明，對於任何正整數 
不知同學會否覺得不太容易？
利用下圖，或可略略帶出式子的意義。先把「垂直」的點數加起，即
再把「水平」的點數加起，即 
利用同樣的圖，考慮不同的數算方法：
不難得出另一個結果：
現在多寫個簡單例子：
考慮 
例如下圖，
長方形 
如用 
![[p][q]](http://s0.wp.com/latex.php?latex=%5Bp%5D%5Bq%5D&bg=ffffff&fg=000000&s=0 "[p][q]")
![[x]](http://s0.wp.com/latex.php?latex=%5Bx%5D&bg=ffffff&fg=000000&s=0 "[x]")
![[5.7]=5](http://s0.wp.com/latex.php?latex=%5B5.7%5D%3D5&bg=ffffff&fg=000000&s=0 "[5.7]=5")
![[7.3]=7](http://s0.wp.com/latex.php?latex=%5B7.3%5D%3D7&bg=ffffff&fg=000000&s=0 "[7.3]=7")
好了，現在用另一個方式數算長方形
引入函數嚴格遞增（strictly increasing）函數 
它的圖像（graph）把格點分成三類：
第一類，在圖像或以下的格點，紅色者：
共有 ![[f(1)]+[f(2)]+\dots +[f(5)]](http://s0.wp.com/latex.php?latex=%5Bf%281%29%5D%2B%5Bf%282%29%5D%2B%5Cdots+%2B%5Bf%285%29%5D&bg=ffffff&fg=000000&s=0 "[f(1)]+[f(2)]+\dots +[f(5)]")
一般地，共有 ![\displaystyle \sum^{[p]}_{k=1}[f(k)]](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csum%5E%7B%5Bp%5D%7D_%7Bk%3D1%7D%5Bf%28k%29%5D&bg=ffffff&fg=000000&s=0 "\displaystyle \sum^{[p]}_{k=1}[f(k)]")
第二類，在圖像或以上的格點，綠色者：
共有 ![[f^{-1}(1)]+[f^{-1}(2)]+\dots +[f^{-1}(7)]](http://s0.wp.com/latex.php?latex=%5Bf%5E%7B-1%7D%281%29%5D%2B%5Bf%5E%7B-1%7D%282%29%5D%2B%5Cdots+%2B%5Bf%5E%7B-1%7D%287%29%5D&bg=ffffff&fg=000000&s=0 "[f^{-1}(1)]+[f^{-1}(2)]+\dots +[f^{-1}(7)]")
一般地，共有 ![\displaystyle \sum^{[q]}_{k=1}[f^{-1}(k)]](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csum%5E%7B%5Bq%5D%7D_%7Bk%3D1%7D%5Bf%5E%7B-1%7D%28k%29%5D&bg=ffffff&fg=000000&s=0 "\displaystyle \sum^{[q]}_{k=1}[f^{-1}(k)]")
第三類，在圖像上的格點，黃色者：
共有 2 個。
所以，如上例，長方形 
一般地，考慮
![\displaystyle \sum^{[p]}_{k=1}[f(k)]+\displaystyle \sum^{[q]}_{k=1}[f^{-1}(k)]-n](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csum%5E%7B%5Bp%5D%7D_%7Bk%3D1%7D%5Bf%28k%29%5D%2B%5Cdisplaystyle+%5Csum%5E%7B%5Bq%5D%7D_%7Bk%3D1%7D%5Bf%5E%7B-1%7D%28k%29%5D-n&bg=ffffff&fg=000000&s=0 "\displaystyle \sum^{[p]}_{k=1}[f(k)]+\displaystyle \sum^{[q]}_{k=1}[f^{-1}(k)]-n")
其中
故推得：
![\displaystyle \sum^{[p]}_{k=1}[f(k)]+\displaystyle \sum^{[q]}_{k=1}[f^{-1}(k)]-n=[p][q]](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csum%5E%7B%5Bp%5D%7D_%7Bk%3D1%7D%5Bf%28k%29%5D%2B%5Cdisplaystyle+%5Csum%5E%7B%5Bq%5D%7D_%7Bk%3D1%7D%5Bf%5E%7B-1%7D%28k%29%5D-n%3D%5Bp%5D%5Bq%5D&bg=ffffff&fg=000000&s=0 "\displaystyle \sum^{[p]}_{k=1}[f(k)]+\displaystyle \sum^{[q]}_{k=1}[f^{-1}(k)]-n=[p][q]")
隨便運用一下 (*)，比如求以式的值：
![[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\dots +[\sqrt{100^2}]](http://s0.wp.com/latex.php?latex=%5B%5Csqrt%7B1%7D%5D%2B%5B%5Csqrt%7B2%7D%5D%2B%5B%5Csqrt%7B3%7D%5D%2B%5Cdots+%2B%5B%5Csqrt%7B100%5E2%7D%5D&bg=ffffff&fg=000000&s=0 "[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\dots +[\sqrt{100^2}]")
解：
設 
則 
代
那麼，在 
由 (*)，得
![([\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\dots +[\sqrt{100^2}])+(1^2+2^2+\dots +100^2)-100=100^2 \times 100](http://s0.wp.com/latex.php?latex=%28%5B%5Csqrt%7B1%7D%5D%2B%5B%5Csqrt%7B2%7D%5D%2B%5B%5Csqrt%7B3%7D%5D%2B%5Cdots+%2B%5B%5Csqrt%7B100%5E2%7D%5D%29%2B%281%5E2%2B2%5E2%2B%5Cdots+%2B100%5E2%29-100%3D100%5E2+%5Ctimes+100&bg=ffffff&fg=000000&s=0 "([\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\dots +[\sqrt{100^2}])+(1^2+2^2+\dots +100^2)-100=100^2 \times 100")
即
一般地，同學試試證明：
![\displaystyle \sum^{n^2}_{k=1}[\sqrt{k}]=\frac{1}{6}n(4n^2-3n+5)](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csum%5E%7Bn%5E2%7D_%7Bk%3D1%7D%5B%5Csqrt%7Bk%7D%5D%3D%5Cfrac%7B1%7D%7B6%7Dn%284n%5E2-3n%2B5%29&bg=ffffff&fg=000000&s=0 "\displaystyle \sum^{n^2}_{k=1}[\sqrt{k}]=\frac{1}{6}n(4n^2-3n+5)")
同學，有信心找 ![\displaystyle \sum^{n^3}_{k=1}[\sqrt[3]{k}]=?](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csum%5E%7Bn%5E3%7D_%7Bk%3D1%7D%5B%5Csqrt%5B3%5D%7Bk%7D%5D%3D%3F&bg=ffffff&fg=000000&s=0 "\displaystyle \sum^{n^3}_{k=1}[\sqrt[3]{k}]=?")
最後，設 
![[\frac{q}{p}]+[\frac{2q}{p}]+[\frac{3q}{p}]+\dots +[\frac{(p-1)q}{p}]=\frac{(p-1)(q-1)}{2}](http://s0.wp.com/latex.php?latex=%5B%5Cfrac%7Bq%7D%7Bp%7D%5D%2B%5B%5Cfrac%7B2q%7D%7Bp%7D%5D%2B%5B%5Cfrac%7B3q%7D%7Bp%7D%5D%2B%5Cdots+%2B%5B%5Cfrac%7B%28p-1%29q%7D%7Bp%7D%5D%3D%5Cfrac%7B%28p-1%29%28q-1%29%7D%7B2%7D&bg=ffffff&fg=000000&s=0 "[\frac{q}{p}]+[\frac{2q}{p}]+[\frac{3q}{p}]+\dots +[\frac{(p-1)q}{p}]=\frac{(p-1)(q-1)}{2}")
同學，試證之。（提示是 
有接觸初等數論的同學，對二次互反律（Law of Quadratic Reciprocity）不會陌生，當中引理的其中一個證明方法，便可利用考慮上述格點數目而得，有興趣也溫習一下吧：
https://en.wikipedia.org/wiki/Quadratic_reciprocity
Also read
點解
https://johnmayhk.wordpress.com/2014/08/09/solve-by-dots/
