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已知数列an中,a1等于1,2nan+1等于(n+1)an,则an的通项公式为

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∵2na(n+1)=(n+1)an,∴a(n+1)/an=(n+1)/(2n),

∴a2/a1=2/(1×2)a3/a2=3/(2×2)a4/a3=4/(2×3)a5/a4=5/(2×4)……an/a(n-1)=n/[2(n-1)]

两边分别相乘,得:an/a1=n/2^(n-1),∵a1=1∴an=n/2^(n-1).

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