来自柯亨玉的问题
已知数列{an}满足an=2a(n-1)+2^n-1(n>或=2),a1=5,bn=(an-1)/(2^n).(1)证明{}为等差数列.(1)证明:{bn}为等差数列
已知数列{an}满足an=2a(n-1)+2^n-1(n>或=2),a1=5,bn=(an-1)/(2^n).(1)证明{}为等差数列.
(1)证明:{bn}为等差数列


已知数列{an}满足an=2a(n-1)+2^n-1(n>或=2),a1=5,bn=(an-1)/(2^n).(1)证明{}为等差数列.(1)证明:{bn}为等差数列
已知数列{an}满足an=2a(n-1)+2^n-1(n>或=2),a1=5,bn=(an-1)/(2^n).(1)证明{}为等差数列.
(1)证明:{bn}为等差数列
an=2a(n-1)+2^(n-1)
两边同除以2^(n+1)
an/2^(n+1)=a(n-1)/2^n+1/4(1)
因bn=a(n-1)/2^n
则b(n+1)=an/2^(n+1)
由(1)知b(n+1)-bn=1/4
所以{bn}是公差为1/4的等差数列