P ( X = k ) = ( n k ) p k q n − k , k = 0 , 1 , 2 , . . , n , q = 1 − p E X = ∑ k = 0 n k ( n k ) p k q n − k = ∑ k = 1 n k ( n k ) p k q n − k = ∑ k = 1 n k n ! k ! ( n − k ) ! p k q n − k = n p ∑ k = 1 n ( n − 1 ) ! ( k − 1 ) ! ( n − k ) ! p k − 1 q ( n − 1 ) − ( k − 1 ) = n p ∑ k = 1 n ( n − 1 k − 1 ) p k − 1 q ( n − 1 ) − ( k − 1 ) = n p [ ( n − 1 0 ) p 0 q n − 1 + ( n − 1 1 ) p 1 q n − 2 + . . . + ( n − 1 n − 1 ) p n − 1 q 0 ] = n p P(X=k) = {n\choose k}p^kq^{n-k}, k = 0,1,2,..,n,q = 1-p\\ EX = \sum_{k=0}^n k {n\choose k}p^kq^{n-k} \\ = \sum_{k=1}^n k {n\choose k}p^kq^{n-k} \\ = \sum_{k=1}^n k {\frac{n!}{k!(n-k)!}}p^kq^{n-k} \\ = np\sum_{k=1}^n {\frac{(n-1)!}{(k-1)!(n-k)!}}p^{k-1}q^{(n-1)-(k-1)} \\ = np\sum_{k=1}^n{n-1\choose k-1}p^{k-1}q^{(n-1)-(k-1)}\\ = np[{n-1\choose 0}p^0q^{n-1}+{n-1\choose 1}p^1q^{n-2}+...+{n-1\choose n-1}p^{n-1}q^0] \\ = np P(X=k)=(kn)pkqn−k,k=0,1,2,..,n,q=1−pEX=k=0∑nk(kn)pkqn−k=k=1∑nk(kn)pkqn−k=k=1∑nkk!(n−k)!n!pkqn−k=npk=1∑n(k−1)!(n−k)!(n−1)!pk−1q(n−1)−(k−1)=npk=1∑n(k−1n−1)pk−1q(n−1)−(k−1)=np[(0n−1)p0qn−1+(1n−1)p1qn−2+...+(n−1n−1)pn−1q0]=np
因为： D X = E X 2 − ( E X ) 2 DX = EX^2-(EX)^2 DX=EX2−(EX)2
且， E X 2 = ∑ k = 1 n k 2 ( n k ) p k q n − k , k = 0 , 1 , 2 , . . , n , q = 1 − p = ∑ k = 1 n [ k ( k − 1 ) + k ] ( n k ) p k q n − k = ∑ k = 1 n k ( k − 1 ) ( n k ) p k q n − k + ∑ k = 1 n k ( n k ) p k q n − k 其 中 ， ∑ k = 1 n k ( n k ) p k q n − k = E X = n p ∑ k = 1 n k ( k − 1 ) ( n k ) p k q n − k = ∑ k = 1 n k ( k − 1 ) n ! k ! ( n − k ) ! p 2 p k − 2 q n − k = ∑ k = 2 n k ( k − 1 ) n ! k ! ( n − k ) ! p 2 p k − 2 q n − k 注 ： 特 别 注 意 这 里 k = 1 时 项 为 0 ， 所 以 可 以 从 k = 2 开 始 计 算 。 = ∑ k = 1 n n ( n − 1 ) ( n − 2 ) ! ( k − 2 ) ! ( n − k ) ! p 2 p k − 2 q [ ( n − 2 ) − ( k − 2 ) ] = n ( n − 1 ) p 2 ∑ k = 2 n ( n − 2 ) ! ( k − 2 ) ! ( n − k ) ! p k − 2 q [ ( n − 2 ) − ( k − 2 ) ] = n ( n − 1 ) p 2 ∑ k = 2 n ( n − 2 k − 2 ) p k − 2 q [ ( n − 2 ) − ( k − 2 ) ] = n ( n − 1 ) p 2 → E X 2 = n ( n − 1 ) p 2 + n p → D X = E X 2 − ( E X ) 2 = n p − n p 2 = n p ( 1 − p ) EX^2 = \sum_{k=1}^nk^2{n\choose k}p^kq^{n-k}, k = 0,1,2,..,n,q = 1-p\\ = \sum_{k=1}^n[k(k-1)+k]{n\choose k}p^kq^{n-k}\\ = \sum_{k=1}^nk(k-1){n\choose k}p^kq^{n-k} + \sum_{k=1}^nk{n\choose k}p^kq^{n-k}\\ 其中， \sum_{k=1}^nk{n\choose k}p^kq^{n-k} = EX = np\\ \sum_{k=1}^nk(k-1){n\choose k}p^kq^{n-k} \\ = \sum_{k=1}^nk(k-1){\frac{n!}{k!(n-k)!}}p^2p^{k-2}q^{n-k} \\ = \sum_{k=2}^nk(k-1){\frac{n!}{k!(n-k)!}}p^2p^{k-2}q^{n-k} \\ 注：特别注意这里k=1时项为0，所以可以从k=2开始计算。\\ = \sum_{k=1}^n{\frac{n(n-1)(n-2)!}{(k-2)!(n-k)!}}p^2p^{k-2}q^{[(n-2)-(k-2)]} \\ = n(n-1)p^2\sum_{k=2}^n{\frac{(n-2)!}{(k-2)!(n-k)!}}p^{k-2}q^{[(n-2)-(k-2)]}\\ = n(n-1)p^2\sum_{k=2}^n{n-2\choose k-2}p^{k-2}q^{[(n-2)-(k-2)]}\\ = n(n-1)p^2 \\ \rightarrow EX^2 = n(n-1)p^2+np \\ \rightarrow DX = EX^2-(EX)^2 = np-np^2 = np(1-p) EX2=k=1∑nk2(kn)pkqn−k,k=0,1,2,..,n,q=1−p=k=1∑n[k(k−1)+k](kn)pkqn−k=k=1∑nk(k−1)(kn)pkqn−k+k=1∑nk(kn)pkqn−k其中，k=1∑nk(kn)pkqn−k=EX=npk=1∑nk(k−1)(kn)pkqn−k=k=1∑nk(k−1)k!(n−k)!n!p2pk−2qn−k=k=2∑nk(k−1)k!(n−k)!n!p2pk−2qn−k注：特别注意这里k=1时项为0，所以可以从k=2开始计算。=k=1∑n(k−2)!(n−k)!n(n−1)(n−2)!p2pk−2q[(n−2)−(k−2)]=n(n−1)p2k=2∑n(k−2)!(n−k)!(n−2)!pk−2q[(n−2)−(k−2)]=n(n−1)p2k=2∑n(k−2n−2)pk−2q[(n−2)−(k−2)]=n(n−1)p2→EX2=n(n−1)p2+np→DX=EX2−(EX)2=np−np2=np(1−p)
核心思想是转化为更小规模的组合数，这里没法直接用幂级数的和函数求解思路。
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