$e^{\ln(2)} + \frac{2 \cdot 2}{2} + \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n + \int_0^1 2 \, dx +\text{tr}\left(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\right) + \left| \{0\} \cup \{1\} \right| + \sum_{k=0}^{1} P(X=k)$

平方差公式推导

$\:\:\:\:\:a^2 - b^2$
$=a[b + (a-b)]-b^2$
$=ab+a(a-b)-b^2$
$=b(a-b)+a(a-b)$
$=(a+b)(a-b)$

四面体数公式推导

令$n = 0,1,2,3,4,...$

$a_n = 0,1, 4, 10, 20,...$

设数列$b$为数列$a$相邻两个元素之差则:

$b_n = 1,3,6,10,...$

再设数列$c$为数列$b$相邻两个元素之差则:

$c_n = 2,3,4,...$

而数列$c$是一个等差数列

$c_n = n+ 2$

由递推关系可得

$c_0 + c_1 + ... + c_n = b_{n-1} - b_0$

所以$b_n=\frac {1}{2}(n+1)(n+2)$

而$b_0 + b_1 + ... + b_n = a_{n+1} - a_0$

$a_{n+1}=1+3+...+\frac{1}{2}n^2+\frac{3}{2}n+1$
$\:\:\:\:\:\:\:\:\:\:=\frac{1}{2}(0+1+4+...+n^2)+\frac{3}{2}(0+1+2+...+n)+(n+1)$
$\:\:\:\:\:\:\:\:\:\:=\frac{1}{12}n(n+1)(2n+1)+\frac{3}{4}(n+1)+(n+1)$
$\:\:\:\:\:\:\:\:\:\:=\frac{1}{12}n(n+1)(2n+1)+\frac{7}{4}(n+1)$

化简得:$\frac{1}{6}n(n+1)(n+2)$