Question

已知数列{an}满足a1+a/4,(1-an)a(n+1)=1/4,令bn+an-1/2

已知数列{an}满足a1+a/4,(1-an)a(n+1)=1/4,令bn+an-1/2 求证数列{1/bn}为等差数列，求和:Sn=a2/a1+a3/a2+...+a(n+1)/an
已知数列{an}满足a1=1/4,(1-an)(a(n+1))=1/4,令bn=an-1/2 求证数列{1/bn}为等差数列，求和:Sn=a2/a1+a3/a2+...+a(n+1)/an

Answer 1

因为(1-an)a(n+1)=1/4 即an*a(n+1)-1/4=0 1/2[(a(n+1)-1/2)-(an-1/2)]-(a(n+1)-1/2)(an-1/2)=0 b(n+1)*bn+1/2(b(n+1)-bn)=0 (b(n+1)-bn)/(b(n+1)*bn)=-2 1/b(n+1)-1/bn=-2 所以{1/bn}为等差数列 1/bn=1/b1-2(n-1)=1/(a1-1/2)+2-2n=-2n-2 bn=-1/(2n+2) an=n/(2(n+1)) a(n+1)/an=(n+1)^2/[n(n+2)]=1+1/2[1/n-1/(n+2)] Sn=n+1/2[1-1/(n+2)]=n+(n+1)/(2n+4)

Answer 2

bn=1+1/n,
Sn=b1+b2+b3+.......+bn
Sn= 1 + 1/1 + 1 + 1/2 + 1 + 1/3 +........+ 1 + 1/n
Sn=n +1 + 1/2 + 1/3 +.......+ 1/n

当n趋于无穷大时，上式可以近似用ln(n) + C来模拟
亦即：
1 + 1/2 + 1/3 + 1/4 + ... + 1/n = ln(n) + C
(其中C为欧拉常数)

Answer 3

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