(1)由程序框图可得:执行循环第一次i=2,f(x)=x+2. 执行循环第二次i=3,f(x)=(x+2)x+3=x2+2x+3 执行循环第三次i=4,f(x)=x3+2x2+3x+4,运行结束. ∴f(x)=x3+2x2+3x+4. (2)f(x)=x10+2x9+3x8+…+10x+11, x10+2x9+3x8+…+10x+11=a10(x-1)10+a9(x-1)9+…a1(x-1)+a0 令x=1,得a0=1+2+…+11==66 f(x)=[(x-1)+1]10+2[(x-1)+1]9+…+10[(x-1)+1]+11 ═a10(x-1)10+a9(x-1)9+…a1(x-1)+a0, ∴a8=C102+2C91+3C80=66. |