设数列an的前n项和为sn,a1=1/4,且2an=2an-1+1,数列bn,满足b1=3/4,且3bn-bn-1=n 求证bn-an为等比数列和bn的通项公式

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设数列an的前n项和为sn,a1=1/4,且2an=2an-1+1,数列bn,满足b1=3/4,且3bn-bn-1=n 求证bn-an为等比数列和bn的通项公式

2an=2a(n-1) +1an = a(n-1) +1/2an -a(n-1) =1/2{an}是等差数列,d=1/2an -a1 = (n-1)/2an = (2n-1)/4 3bn-b(n-1)=nbn -(1/2)n + 1/4= (1/3)[ b(n-1) -(1/2)(n-1) + 1/4]{bn -(1/2)n + 1/4} 是等比数列,q= 1/3bn -(1/...