MATH

组合数学：
\[\sum\limits_{k = 1}^n {{{(2k - 1)}^2} = \frac{{n(4{n^2} - 1)}}{3}}\]

\[\sum\limits_{k = 1}^n {{{(2k - 1)}^3} = {n^2}(2{n^2} - 1)}\]

\[{\sum\limits_{k = 1}^n {{k^3} = \left( {\frac{{n(n + 1)}}{2}} \right)} ^2}\]

\[\sum\limits_{k = 1}^n {{k^4} = \frac{{n(n + 1)(2n + 1)(3{n^2} + 3n - 1)}}{{30}}}\]

\[\sum\limits_{k = 1}^n {{k^5} = \frac{{{n^2}{{(n + 1)}^2}(2{n^2} + 2n - 1)}}{{12}}}\]

\[\sum\limits_{k = 1}^n {k(k + 1) = \frac{{n(n + 1)(n + 2)}}{3}}\]

\[\sum\limits_{k = 1}^n {k(k + 1)(k + 2) = \frac{{n(n + 1)(n + 2)(n + 3)}}{4}}\]

\[\sum\limits_{k = 1}^n {k(k + 1)(k + 2)(k + 3) = \frac{{n(n + 1)(n + 2)(n + 3)(n + 4)}}{5}} \]

数论公式：

\[\mathop {\lim }\limits_{n \to + \infty } \frac{{\pi (n)}}{{n/\ln n}} = 1\]

\[\ln n - \frac{3}{2} \le \frac{n}{{\pi (n)}} \le \ln n - \frac{1}{2}\left( {n \ge 67} \right)\]

\[n! \approx \sqrt {2\pi n} {\left( {\frac{n}{e}} \right)^n} \]

\[({a^m} - 1,{a^n} - 1) = {a^{(m,n)}} - 1\left( {a > 1,m,n > 0} \right) \]

\[({a^m} - {b^m},{a^n} - {b^n}) = {a^{(m,n)}} - {b^{(m,n)}}\left( {a > b,\gcd (a,b) = 1} \right) \]

\[({F_n},{F_m}) = {F_{(n,m)}}\left( {{F_n} = {F_{n - 1}} + {F_{n - 2}}} \right) \]

\[\sum\limits_{i = 1}^N {\gcd (i,N) = \sum\limits_{d|N} {d\varphi (N/d)} } \]

\[\sum\limits_{i = 1}^N {\frac{N}{{\gcd (i,N)}} = \sum\limits_{d|N} {d\varphi (d)} } = (\frac{{{p_1}^{2{a_1} + 1} + 1}}{{{p_1} + 1}})(\frac{{{p_2}^{2{a_2} + 1} + 1}}{{{p_2} + 1}}) \times ... \times (\frac{{{p_k}^{2{a_k} + 1} + 1}}{{{p_k} + 1}})\left( {N = {p_1}^{{a_1}}{p_2}^{{a_2}}...{p_k}^{{a_k}}} \right) \]

\[(n + 1)lcm(C_n^0,C_n^1,...C_n^{n - 1},C_n^n) = lcm(1,2,...n + 1) \]

\[\gcd (ab,m) = \gcd (a,m) \times \gcd (b,m) \]

转载于:https://www.cnblogs.com/mj-liylho/p/7679725.html