解(3)f(x) (x+1)(x+2)x+ 其中 12+(x-1) X-1 l<x<3 ∑(-1 x+23+(x-1)3 x+1x+2 (( n+2n+1x-1 1<X<3 2x2+4x+12 如f(x)=h(-x+x2) =hn(+x3)-hn(+x) 1+X ∑(1)1(k2n-x)(-1, n 解(4)f(x)= 1+x2+(x-1)2 其中 x"|=2∑ ∑x2=2∑n+1)x2-∑解⑶ ( ) ( )( ) x 2 1 x 1 1 x 1 x 2 1 f x + − + = + + = 其中 ( ) ( )       − = −       − + = + − = +  n=0 n n 2 x 1 1 2 1 2 x 1 2 1 1 2 x 1 1 x 1 1 1 2 x 1 1  − −  ，−1 x  3 ( )  ( )       − = −       − + =  + − = +  n=0 n n 3 x 1 1 3 1 3 x 1 1 1 3 1 3 x 1 1 x 2 1 − 2  x  4 ∴ ( )  ( ) ( − )      = − − + − + =  = + + n 0 n n 1 n 1 n x 1 3 1 2 1 1 x 2 1 x 1 1 f x −1 x  3 （如 x0 = 0 ， ( ) 2 x 1 1 2 1 1 x 1 f x + −  + = ( ) 2x 3 1 5 1 x 4 1 5 2 2x 4x 12 x 2 f x 2 + +  + =  + + + = 如 ( ) ( ) ln (1 x ) ln(1 x) 1 x 1 x f x ln 1 x x ln 3 3 2 = + − +         + + = − + = =  (− ) ( − )  = − n 1 n 1 3n n x x n 1 1 (−1 ，１ ） 解⑷ ( ) ( ) ( ) ( ) ( ) 1 x 1 1 x 2 1 x 2 x 1 1 x 1 x f x 2 2 2 − − − = − + − = − + = 其中 ( ) ( ) =         =           − = −  = −  = n 1 n 1 n 0 n 2 2 x 2 nx 1 x 2 1 x 2 ( ) =  −  = ( + ) −  = ( + )  =  =  =  =  = − n 0 n 0 n n 0 n n n 0 n n 1 n 1 f x 2 nx x 2 n 1 x x 2n 1 x −1 x 1