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数列an满足a1=2,an+1(n+1为下标)-=2n+1n+1为上标)*an/(n+1/2)an+2的n次方
(1)设bn=2的n次方除以an,求数列{bn}的通项公式
(2)设cn=1/n(n+1)an+1(n+1为下标)
人气:337 ℃ 时间:2019-08-16 23:54:05
解答
你问的是不是这样?a(n+1)=2^(n+1)an/[(n+1/2)an+2^n](1)易得2^(n+1)/a(n+1)=[(n+1/2)an+2^n]/an=n+1/2+2^n/an即b(n+1)=bn+n+1/2bn=b(n-1)+(n-1)+1/2=b1+1+2+……+(n-1)+1/2(n-1)=1+1/2[n(n-1)]+1/2(n-1)=1/2(n^2+1)(...
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