设函数f(x)=a1+a2x+a3x^2+a4x^3+.+anx^n-1,f(0)=1/2,数列{an}满足f(1)=n^2an,（n>=1）,求数列{an}

Question

设函数f(x)=a1+a2x+a3x^2+a4x^3+.+anx^n-1,f(0)=1/2,数列{an}满足f(1)=n^2an,（n>=1）,求数列{an}

Answer 1

∵函数f（x）=a1+a2x+a3x2+…+anxn-1,f（0）=1/2 ,
∴a1=1/2 ,
∵f（1）=n2•an,
∴a1+a2+a3+…+an=n2•an,
又∵an=Sn-Sn=n2•an-（n-1）2•an-1,
∴（n2-1）an=（n-1）2•an-1（n≥2）,
an/an-1 =n2-1/(n-1)2 =n+1/n-1
a2/a1 •a3/ a2 •a4 /a3 …an /an-1 =1/3 ×2/4 ×…×n-2/n ×n-1/n+1 ,
an/an-1 =1/3 ×2/4 ×…×n-2/n ×n-1/n+1 ,
∴an=1/n(n+1) ,