已知数列{an}中,an=1+1/2+1/3+...+1/n,记sn=a1+a2+...+an用数学归纳法证明sn=(n+1)an-n
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![已知数列{an}中,an=1+1/2+1/3+...+1/n,记sn=a1+a2+...+an用数学归纳法证明sn=(n+1)an-n](/uploads/image/z/5457271-31-1.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E4%B8%AD%2Can%3D1%2B1%2F2%2B1%2F3%2B...%2B1%2Fn%2C%E8%AE%B0sn%3Da1%2Ba2%2B...%2Ban%E7%94%A8%E6%95%B0%E5%AD%A6%E5%BD%92%E7%BA%B3%E6%B3%95%E8%AF%81%E6%98%8Esn%3D%28n%2B1%29an-n)
已知数列{an}中,an=1+1/2+1/3+...+1/n,记sn=a1+a2+...+an用数学归纳法证明sn=(n+1)an-n
已知数列{an}中,an=1+1/2+1/3+...+1/n,记sn=a1+a2+...+an用数学归纳法证明sn=(n+1)an-n
已知数列{an}中,an=1+1/2+1/3+...+1/n,记sn=a1+a2+...+an用数学归纳法证明sn=(n+1)an-n
(1) 当n=1时
S1=(1+1)a1-1
=2a1-1
=2-1
=1
∴当n=1时sn=(n+1)an-n成立
(2)假设n=k时sn=(n+1)an-n成立
即有 Sk=(k+1)ak-k
Sk+1=1+1/2+1/3+...+1/k+1/(k+1)
=Sk+1/(k+1)
=(k+1)ak-k+1/(k+1)
=(k+1)/k-k+1/(k+1)
=1+1/k-k+1/(k+1)
=(k+2)/(k+1)-(k+1)
=(k+2)ak+1-(k+1)
即当n=k+1时 Sk+1=(k+2)ak+1-(k+1) 成立
综上所述,当n∈N* 时有sn=(n+1)an-n 成立
(i)当n=1时,左边=a1=1,右边=(1+1)a1-1=1 等式成立
(ii)假设当n=k(k属于N*,k>=1)时等式成立
即 sk=(k+1)ak-k
当n=k+1时 ak+1=ak+1/(1+k)
sk+1=sk+ak+1=(k+1)ak-k+ak+1
=(k+1...
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(i)当n=1时,左边=a1=1,右边=(1+1)a1-1=1 等式成立
(ii)假设当n=k(k属于N*,k>=1)时等式成立
即 sk=(k+1)ak-k
当n=k+1时 ak+1=ak+1/(1+k)
sk+1=sk+ak+1=(k+1)ak-k+ak+1
=(k+1+1)ak+1-k-(k+1)/(k+1)
=(k+1+1)ak+1-k-1
等式也成立
根据(i)(ii)可以断定sn=(n+1)an-n对于n属于N*都成立
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