# Quod Erat Demonstrandum

## 2016/08/08

### 點解2

Filed under: Fun,mathematics — johnmayhk @ 11:12 下午
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$1+2+3+\dots +(n-1)+n+(n-1)+\dots +3+2+1=?$

$\frac{n(n+1)}{2}+\frac{(n-1)n}{2}=n^2$

$1+2+3+\dots +(n-1)+n+(n-1)+\dots +3+2+1=n^2$

$1+2+3+4+5+6+5+4+3+2+1=6^2$

$[5.7]=5$
$[7.3]=7$

$\displaystyle \sum^{[p]}_{k=1}[f(k)]+\displaystyle \sum^{[q]}_{k=1}[f^{-1}(k)]-n$

$\displaystyle \sum^{[p]}_{k=1}[f(k)]+\displaystyle \sum^{[q]}_{k=1}[f^{-1}(k)]-n=[p][q]$ ……… (*)

$[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\dots +[\sqrt{100^2}]$

$f(x)=\sqrt{x}$

$f^{-1}(x)=x^2$（為何？）

$p=100^2,q=100$

$n=100$

$([\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\dots +[\sqrt{100^2}])+(1^2+2^2+\dots +100^2)-100=100^2 \times 100$

$[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\dots +[\sqrt{100^2}]$

$=100^3+100-\frac{1}{6}(100)(100+1)(200+1)$　　（因為 $1^2+2^2+\dots +n^2=\frac{1}{6}n(n+1)(2n+1)$

$=661750$

$\displaystyle \sum^{n^2}_{k=1}[\sqrt{k}]=\frac{1}{6}n(4n^2-3n+5)$

$[\frac{q}{p}]+[\frac{2q}{p}]+[\frac{3q}{p}]+\dots +[\frac{(p-1)q}{p}]=\frac{(p-1)(q-1)}{2}$