X+x+.+x X =[ [(2x)°+(2x)4+…+(2x)n+…]-[x0+x2+…+x+…]* [(21-1)x2+…+(2”-1)x”]*[x+x1+…+x”+… 结果中,x"的系数为: (21-1)+(22-1)+…+(21-1)+(2n-1) 2(1-2") 1-2 此即T(n)=2n+-2 另 令G1(x)= 则 G()=(G1(x)y=(△Jkdy= jx=xG,(x) ②展开法: T(n)=2T(n-1)+n =2(2T(n-2)+n-1)+n=22(n-2)+2(n-1)+n 22(2(n-3)+n-2)+2(m-1)+n 23T(n-3)+22(n-2)+2(n-1) 21(1)+2”-2*2+2-3*3+…+20米n =2*1+2 +…+2*n= 1 x x x ... x ... 2 n - + + + + = 1 x 1 x x - - = 2 (1 x) x - G(x)= 1 2x 1 - * 1 x x - * 1 x 1 - =[ 1 2x 1 - - 1 x x - ][ 1 x 1 - ] =[(2x) 0 + (2x)1 +…+(2x) n +…]-[x 0 +x1 +…+x n +…]]* 1- x 1 [(21 -1)x1 +…+(2 n -1)x n ]*[ x 0 +x1 +…+x n +…] 结果中，x n 的系数为： (21 -1)+( 2 2 -1)+…+(2 n-1 -1)+ (2 n -1) =21 + 2 2 +…+2 n -n= 1 2 2(1 2 ) - - n -n=2 n+1` -2-n 此即 T(n)= 2 n+1` -2-n 另：å ¥ j=1 j jx =xå ¥ = - j 1 j 1 jx 令 G1 (x)= å ¥ = - j 1 j 1 jx 则 G1(x)= ò ( G (x)dx)¢ 1 = ( jx dx) j 1 j 1 å ¢ ò ¥ = - = 2 (1 x ) 1 - =>å ¥ j=1 j jx =x G1 (x)= 2 (1 x ) x - ②展开法： T(n)=2T(n-1)+n =2(2T(n-2)+n-1)+n=2 2 T(n-2)+2(n-1)+n =2 2 (2T(n-3)+n-2)+2(n-1)+n =2 3 T(n-3)+ 2 2 (n-2)+2(n-1)+n M =2 n-1 T(1)+ 2 n-2 *2+2 n-3 *3+…+2 0 *n =2 n-1 *1+2 n-2 *2+…+2 0 *n