数列 (28 12:41:33)

数列 (28 12:41:33)
 
已知数列{an},{bn}满足bn+1=bn+4an,bn=(2n-1)an
求证：数列{an}为等差数列
若a2=a1+b1-1,求∑(1/b)
(要完整过程)

(n+1)=bn+4an,bn=(2n-1)an
所以:(2n+1)a(n+1)=(2n-1)an+4an=(2n+3)an
所以:an=((2n+1)/(2n-1))a(n-1)
=((2n+1)/(2n-1))((2n-1)/(2n-3))a(n-2)
=((2n+1)/(2n-1))((2n-1)/(2n-3))((2n-3)/(2n-5))a(n-3)
=...
=((2n+1)/(2n-1))((2n-1)/(2n-3))((2n-3)/(2n-5))...(7/5)(5/3)a1
=((2n+1)/3)a1
所以:a(n+1)-an=(2/3)a1,
数列{an}为等差数列
a2=a1+b1-1
所以:b1-1=(2/3)a1
b1=1+(2/3)a1
而由:bn=(2n-1)an,则:b1=a1
所以:a1=1+(2/3)a1
a1=3
而:an=((2n+1)/3)a1=2n+1
bn=(2n-1)an=(2n-1)(2n+1)
1/bn=1/((2n-1)(2n+1))
=(1/2)((1/(2n-1))-(1/(2n+1)))
∑(1/b)=(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)