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  • MATH

    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) ]

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  • 原文地址:https://www.cnblogs.com/mj-liylho/p/7679725.html
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