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 imes q^{k-1}=egin{cases} an\,\,\,\,\,\,\,\,\, q=1\{{a(1-q^n)} over {1-q}} \,\,\,\,\, q
ot= 1 \ end{cases}\
S_n=sum_{k=1}^n a imes q^{k-1}\
qS_n=sum_{k=1}^n a imes q^k\
(1-q)S_n=a-a imes 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 {nchoose k}x^{n-k}y^k
]