常用数学知识
组合数的一些性质
\[C_n^m=\frac{n!}{(n-m)!m!}\\
C_n^m=C_n^{n-m}\\
C_n^m=C_{n-1}^{m-1}+C_{n-1}^m\\
C_{m+r+1}^{r}=\sum_{i=0}^rC_{m+i}^i\\
\]
\[C_n^mC_m^r=\frac{n!}{m!(n-m)!}.\frac{m!}{r!(m-r)!}\\
C_n^mC_m^r=\frac{n!}{(n-m)!r!(m-r)!}\\
C_n^mC_m^r=\frac{n!(n-r)!}{r!(n-m)!(m-r)!(n-r)!}\\
C_n^mC_m^r=\frac{n!}{r!(n-r)!}\frac{(n-r)!}{(n-m)!(m-r)!}\\
n-m=n-r-(m-r)\\
C_n^mC_m^r=C_n^rC_{n-r}^{m-r}\\
\]
\[\sum_{i=0}^nC_n^i=2^n
\]
\(C_n^i\)可以看做n位二进制数有x个0的数的方案数，推广一下:
\[\sum_{i=0}^nC_n^ix^{n-i}=(x+1)^n
\]
\[\sum_{k=1}^nk^2={{k(k+1)(2k+1)}\over 6}\\
使用数学归纳法证明\\
f(1)=1={{1(1+1)(2+1)}\over{6}}\\
f(k)=\sum_{k=1}^nk^2={{k(k+1)(2k+1)}\over 6}\\
f(k+1)=f(k)+(k+1)^2\\
f(k+1)={{k(k+1)(2k+1)}\over 6}+{{6(k+1)(k+1)}\over{6}}\\
f(k+1)={{(k+1)[k(2k+1)+6(k+1)]}\over 6}\\
f(k+1)={{(k+1)(2k^2+7k+6)}\over 6}\\
f(k+1)={{(k+1)(k+2)(2k+3)}\over{6}}\\
f(k+1)={{k+1[(k+1)+1][2(k+1)+1]} \over 6}\\
\]
\[\sum_{k=1}^n a\times q^{k-1}=\begin{cases}\ an\,\,\,\,\,\,\,\,\, q=1\\{{a(1-q^n)} \over {1-q}} \,\,\,\,\, q\not= 1 \\ \end{cases}\\
S_n=\sum_{k=1}^n a\times q^{k-1}\\
qS_n=\sum_{k=1}^n a\times q^k\\
(1-q)S_n=a-a\times q^n\\
S_n={{a(1-q^n)}\over{1-q}}
\]
\[\sum_{i=1}^n i^3=[{{n(n+1)}\over{2}}]^2\\
数学归纳法证明\\
f(1)=[{{1(1+1)}\over{2}}]^2=1\\
f(n)=\sum_{i=1}^n i^3=[{{n(n+1)}\over{2}}]^2\\
f(n+1)=f(n)+(n+1)^3\\
f(n+1)=[{{n(n+1)}\over{2}}]^2+(n+1)^3\\
f(n+1)={{n^2(n+1)^2+4(n+1)^3}\over 4}\\
f(n+1)={{(n+1)^2(n^2+4n+4)}\over 4}\\
f(n+1)={{(n+1)^2(n+2)^2}\over 4}\\
f(n+1)=[{{(n+1)(n+2)}\over{2}}]^2\\
\]
\[(x+y)^n=\sum_{k=0}^n {n\choose k}x^{n-k}y^k
\]
