已知数列{an}满足a1=1/2,a1+a2+……+an=n^2an(n∈N*),试用数学归纳法证明:an=1/[n(n+1)]好的分再加= =
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![已知数列{an}满足a1=1/2,a1+a2+……+an=n^2an(n∈N*),试用数学归纳法证明:an=1/[n(n+1)]好的分再加= =](/uploads/image/z/1755649-1-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%2F2%2Ca1%2Ba2%2B%E2%80%A6%E2%80%A6%2Ban%3Dn%5E2an%28n%E2%88%88N%2A%29%2C%E8%AF%95%E7%94%A8%E6%95%B0%E5%AD%A6%E5%BD%92%E7%BA%B3%E6%B3%95%E8%AF%81%E6%98%8E%3Aan%3D1%2F%5Bn%28n%2B1%29%5D%E5%A5%BD%E7%9A%84%E5%88%86%E5%86%8D%E5%8A%A0%3D+%3D)
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已知数列{an}满足a1=1/2,a1+a2+……+an=n^2an(n∈N*),试用数学归纳法证明:an=1/[n(n+1)]好的分再加= =
已知数列{an}满足a1=1/2,a1+a2+……+an=n^2an(n∈N*),试用数学归纳法证明:an=1/[n(n+1)]
好的分再加= =
已知数列{an}满足a1=1/2,a1+a2+……+an=n^2an(n∈N*),试用数学归纳法证明:an=1/[n(n+1)]好的分再加= =
当n=1时显然成立.
若当n=m时也成立(m>=1),则当n=m+1时有
m^2am+a(m+1)=(m+1)^2a(m+1) =>m^2/[m(m+1)]=(m^2+2m)a(m+1)
=>a(m+1)=m^2/[(m^2+m)(m^2+2m)]
=>a(m+1)=1/[(m+1)(m+2)]
=>a(m+1)=1/{(m+1)[(m+1)+1)]}
得证!
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