设数列{bn}满足b1=3,bn=3^nP^n,且Pn+1=Pn+n/3^n+1,若存在实数t,使得数列Cn=[bn-(1/4)]*t/(n+1)+n成等差数列,记数列{Cn*(1/2)^Cn}的前n项和为Tn.证明:3^n*(Tn-1)<

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设数列{bn}满足b1=3,bn=3^nP^n,且Pn+1=Pn+n/3^n+1,若存在实数t,使得数列Cn=[bn-(1/4)]*t/(n+1)+n成等差数列,记数列{Cn*(1/2)^Cn}的前n项和为Tn.证明:3^n*(Tn-1)<bn

我就提提思路,要是全打出来太费劲了先求Pn仿写P(n+1)=Pn+n/3^(n+1) Pn=Pn-1+n-1/3^n .P3=P2+(3-1)/3^3P2=P1+(2-1)/3^ 到此为止全相加,消去不少最后得P(n+1)=P1+{数列(n-1)/3^n的第二项...