已知数列{an}满足a1=1,Sn=[(n+1)/2]*an,求通项an.急,思路也要
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![已知数列{an}满足a1=1,Sn=[(n+1)/2]*an,求通项an.急,思路也要](/uploads/image/z/7252448-32-8.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%2CSn%3D%5B%28n%2B1%29%2F2%5D%2Aan%2C%E6%B1%82%E9%80%9A%E9%A1%B9an.%E6%80%A5%2C%E6%80%9D%E8%B7%AF%E4%B9%9F%E8%A6%81)
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已知数列{an}满足a1=1,Sn=[(n+1)/2]*an,求通项an.急,思路也要
已知数列{an}满足a1=1,Sn=[(n+1)/2]*an,求通项an.急,思路也要
已知数列{an}满足a1=1,Sn=[(n+1)/2]*an,求通项an.急,思路也要
由题意得an=Sn-S(n-1)这个你知道吧?n-1是Sn下面的
∴an=[(n+1)/2]*an-[(n-1+1)/2]*a(n-1) 这个n-1也是 an下面的
∵d=an-a(n-1)∴an=nd你自己总能化简吧?
然后代∵an=a1+(n-1)d
把an=nd代入 得d=1
∴an=1+(n-1)*1
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