()=∑k(k-1)pk= k-2 k=1 ● ∑k2 k-2 ∑kpz k-2 k=1 k=1 g"(1)=E(x)-E(X)=E(X)-g( E(X2)=g"(l)+g( 即D(X)=g(1)+g(1)-[g(1)2  = −  = − 1 2 ( ) ( 1) k k k  g z k k p z    =  = − − = − 1 1 2 2 2 k k k k k k k p z k p z (1) ( ) ( ) 2 g  = E X − E X ( ) (1) 2 = E X − g  ( ) (1) (1) 2 E X = g  + g  即 2 D(X) = g (1) + g (1) −[g (1)]