Question

(1)在数列{an},a1=2,an=2an-1+2n+1(n≥2,n∈N*),令{bn}= an/2n,求证：{bn}

Answer 1

1,an=2a(n-1)+2^(n+1),

两边除以2^n,得

an/2^n=a(n-1)/2^(n-1)+2,

即bn=b(n-1)+2,

所以{ bn }是等差数列,

b1=a1/2=1,bn=1+(n-1)*2=2n-1；

2,1/bnb(n+1)=1/(2n-1)(2n+1)=1/2*[1/(2n-1)-1/(2n+1)],

所以Tn=1/b1b2 +1/ b2b3 +.+1/ bnb（n+1）

=1/2*[(1/1-1/3) +(1/3-1/5)+.+1/(2n-1)-1/(2n+1)]

=1/2*(1-1/(2n+1)])

=n/(2n+1).

b[1]=a[1]/2=1,所以b[n]=1+2(n-1)=2n-1

因为b[n]=b[n-1]+2,所以1/(b[n]b[n-1])=(1/b[n-1]-1/b[n])/2,那么

T[n]=(1/b[1]-1/b[2])/2+…+(1/b[n-1]-1/b[n])/2

=(1/b[1]-1/b[n])/2

=(n-1)/(2n-1)