已知数列a1=3 an*an-1=2an-1 -1 求2a2.a3.a4 求证1/an-1是等差数列 并写出an的通项

已知数列a1=3 an*an-1=2an-1 -1 求2a2.a3.a4 求证1/an-1是等差数列 并写出an的通项
已知数列a1=3 an*an-1=2an-1 -1 求2a2.a3.a4 求证1/an-1是等差数列 并写出an的通项

已知数列a1=3 an*an-1=2an-1 -1 求2a2.a3.a4 求证1/an-1是等差数列 并写出an的通项
an*a(n-1)=2a(n-1)-1
an=[2a(n-1)-1]/a(n-1)
an=2-1/a(n-1)
an -1= 1-1/a(n-1)= [a(n-1)-1]/a(n-1)
1/(an -1) = a(n-1)/[a(n-1)-1] = {[a(n-1)-1]+1} /[a(n-1)-1]
=1+ 1/[a(n-1) -1]
所以,1/(an -1) -1/[a(n-1) -1] =1
所以{1/(an-1)} 是等差数列.首项1/(a1-1)=1/2,公差是1
1/(an-1)=1/2+(n-1)*1=n+1/2
an-1=1/(n+1/2)=2/(2n+1)
an=1+2/(2n+1)=(2n+3)/(2n+1)
而a1=(2+3)/3=5/3不=3.
所以,通项是:
a1=3,(n=1)
an=(2n+3)/(2n+1),(n>=2)