已知数列(An)满足A1=1,An=(4A(n-1))/(kA(n-1)+1)(n>=2) (1)求数列An的通项公式
问题描述:
已知数列(An)满足A1=1,An=(4A(n-1))/(kA(n-1)+1)(n>=2) (1)求数列An的通项公式
答
科学归纳法:
A1=1
A2=4A1/(kA1+1)=4/(k+1)=4^(2-1)/(4^0*k+1)
A3=4A2/(kA2+1)=16/(k+1)/[4k/(k+1)+1]=16/(5k+1)=4^(3-1)/[(4^0+4^1)k+]
A4=4A3/(kA3+1)=4*16/(5k+1)/[16k/(5k+1)+1]=64/(21k+1)=4^(4-1)/[(4^0+4^1+4^2)k+1]
.
An=4^(n-1)/{[4^(n-1)-1]/3+1} (n>=2)
n=1时,上式也成立
An=4^(n-1)/{[4^(n-1)-1]/3+1}