组合数学：
[sumlimits_{k = 1}^n {{{(2k - 1)}^2} = frac{{n(4{n^2} - 1)}}{3}}
]
[sumlimits_{k = 1}^n {{{(2k - 1)}^3} = {n^2}(2{n^2} - 1)}
]
[{sumlimits_{k = 1}^n {{k^3} = left( {frac{{n(n + 1)}}{2}}
ight)} ^2}
]
[sumlimits_{k = 1}^n {{k^4} = frac{{n(n + 1)(2n + 1)(3{n^2} + 3n - 1)}}{{30}}}
]
[sumlimits_{k = 1}^n {{k^5} = frac{{{n^2}{{(n + 1)}^2}(2{n^2} + 2n - 1)}}{{12}}}
]
[sumlimits_{k = 1}^n {k(k + 1) = frac{{n(n + 1)(n + 2)}}{3}}
]
[sumlimits_{k = 1}^n {k(k + 1)(k + 2) = frac{{n(n + 1)(n + 2)(n + 3)}}{4}}
]
[sumlimits_{k = 1}^n {k(k + 1)(k + 2)(k + 3) = frac{{n(n + 1)(n + 2)(n + 3)(n + 4)}}{5}}
]
数论公式：
[mathop {lim }limits_{n o + infty } frac{{pi (n)}}{{n/ln n}} = 1
]
[ln n - frac{3}{2} le frac{n}{{pi (n)}} le ln n - frac{1}{2}left( {n ge 67}
ight)
]
[n! approx sqrt {2pi n} {left( {frac{n}{e}}
ight)^n}
]
[({a^m} - 1,{a^n} - 1) = {a^{(m,n)}} - 1left( {a > 1,m,n > 0}
ight)
]
[({a^m} - {b^m},{a^n} - {b^n}) = {a^{(m,n)}} - {b^{(m,n)}}left( {a > b,gcd (a,b) = 1}
ight)
]
[({F_n},{F_m}) = {F_{(n,m)}}left( {{F_n} = {F_{n - 1}} + {F_{n - 2}}}
ight)
]
[sumlimits_{i = 1}^N {gcd (i,N) = sumlimits_{d|N} {dvarphi (N/d)} }
]
[sumlimits_{i = 1}^N {frac{N}{{gcd (i,N)}} = sumlimits_{d|N} {dvarphi (d)} } = (frac{{{p_1}^{2{a_1} + 1} + 1}}{{{p_1} + 1}})(frac{{{p_2}^{2{a_2} + 1} + 1}}{{{p_2} + 1}}) imes ... imes (frac{{{p_k}^{2{a_k} + 1} + 1}}{{{p_k} + 1}})left( {N = {p_1}^{{a_1}}{p_2}^{{a_2}}...{p_k}^{{a_k}}}
ight)
]
[(n + 1)lcm(C_n^0,C_n^1,...C_n^{n - 1},C_n^n) = lcm(1,2,...n + 1)
]
[gcd (ab,m) = gcd (a,m) imes gcd (b,m)
]