设数列an与数列bn满足a1=b1=1,bn/an=1/a1 +1/a2+……+1/an-1,求证:设(1+1/b1)(1+1/b2)……(1+1/bn)=入(1/a1+1/a2+……1/an).求实数入的值老师说用数学归纳法,怎么证明

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设数列an与数列bn满足a1=b1=1,bn/an=1/a1 +1/a2+……+1/an-1,求证:设(1+1/b1)(1+1/b2)……(1+1/bn)
=入(1/a1+1/a2+……1/an).求实数入的值
老师说用数学归纳法,怎么证明

首先要先确定出λ的值.n=1,则1+1/b1=λ/a1∴λ=2
数学归纳法证明如下:
当λ=1时,成立;
假设当λ=k时,成立.即(1+1/b1)(1+1/b2)……(1+1/bk)=2(1/a1+1/a2+……1/ak).
∴(1+1/b1)(1+1/b2)……(1+1/bk)(1+1/b)=2(1/a1+1/a2+...+1/ak)(1+1/b)=2[(1/a1 +...+1/ak)+(1/a1+1/a2+...+1/ak)*(1/b)]=2[(1/a1+1/a2+...+1/ak)+ ( b/a)*(1/b)]=2[1/a1+1/a2+...+1/ak+1/a],故得证
综上,结论成立
说明:b表示数列{bn}的第k+1项

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