1/an=1/4*1/a(n-1)+k/4
1/an-k/3=1/4[1/a(n-1)-k/3]
当k=3时,1/a1-1=01/an-1=0,an=1
当k不等于3时,｛1/an-k/3｝是以1/a1-k/3=1-k/3为首项,1/4为公比的等比数列
1/an-k/3=(1-k/3)(1/4)^(n-1)
an=3*4^(n-1)/[k*4^(n-1)+3-k]
an-3/k=(3k-9)/[k^2*4^(n-1)+k(3-k)]>(3k-9)/k^2*1/4^(n-1)
Sn-(3n-8k)/k=(a1-3/k)+(a2-3/k)+...(an-3/k)+8
>(3k-9)/k^2*[1+1/4+1/4^2+...+1/4^(n-1)]+8
=4(k-3)/k^2*[1-(1/4)^n]+8
>4(k-3)/k^2+8=4(2k+3)(k-1)/k^2>0
所以Sn>(3n-8k)/k
请不要出错题,算了半天都不对,一验证,发现结论反过来了