若（2x+1）4 =a0x4 +a1x3 +a2x2 +a3x+a4,试求：（1）a0+a1+a2+a3+a4的值.（2

（1）把x = 1代入上式可得：f(1) = (2 +1) 4 = a 0 *1 + a 1 *1 + a 2 *1 + a 3 *1+ a 4 = a 0 + a 1 + a 2 + a 3 + a 4 ,所以a 0 + a 1 + a 2 + a 3 + a 4 = 3 4 = 81 ①； （2）把x = -1代入解析式可得：f(-1) = (-2+ 1) 4 = a 0 *(-1) 4 + a 1 *(-1) 3 + a 2 *(-1) 2 + a 3 *(-1) + a 4 = a 0 –a 1 + a 2 –a 3 + a 4 ,所以a 0 –a 1 + a 2 –a 3 + a 4 = (-1) 4 = 1 ②； 把①+②,可得2a 0 + 2a 2 + 2a 4 = 81 + 1 = 82,所以a 0 + a 2 + a 4 = 41 ； 综上所述,a 0 + a 1 + a 2 + a 3 + a 4 = 81 ,a 0 + a 2 + a 4 = 41 .