Question

求教一条高一数学题,

已知数列｛an｝(n为底数)是等比数列,数列｛bn｝的前n项和为sn(n为底数),且a1=1,a4=125/64,bn＞0,4sn=bn^2+2bn+1(n属于正整数)证明存在K属于正整数,,使得Sn/an≤Sk/ak(这里的k为底数)对任意n属于正整数的均成立.请写详细过程

Answer 1

a4=a1q^3q=5/4an=(5/4)^(n-1)4bn=4Sn-4Sn-1=(bn^2+2bn+1)-(bn-1^2+2bn-1+1)bn^2-bn-1^2=2(bn+bn-1)bn-bn-1=24b1=4S1=b1^2+2b1+1b1=1bn=1+2(n-1)=2n-1Sn=[(bn+1)/2]^2=n^2设y(n)=Sn/an=n^2*(4/5)^(n-1)y'(n)=2n*(4/5)^(n-1)+n^2*(4/5)^(n-1)*ln(4/5)=(4/5)^(n-1)*[2n-n^2*ln(5/4)]=0n=2/ln(5/4)≈8.96所以，当0<n≤8时，y'(n)>0，Sn/an是单调递增，当n≥9时，y'(n)<0，Sn/an是单调递减S8/a8-S9/a9=(4/5)^7*(64-81*4/5)<0S8/a8<S9/a9取k=9，则对任意n(n∈R+)，都有Sn/an≤Sk/ak成立