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）
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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）
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）
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）
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+1/2
bn=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)
(2)an=2^n/bn=2^(n+1)/(n^2+1)
cn=(n^2+2n+2)/[2^(n+2)(n^2+n)]=(n+1)/[2^(n+1)n]-(n+2)/[2^(n+2)(n+1)]
前n项和为Sn=(2/4*1-3/8*2)+(3/8*2-4/16*3)+……+{(n+1)/[2^(n+1)n]-(n+2)/[2^(n+2)(n+1)]}
=2/4*1-(n+2)/[2^(n+2)(n+1)]（裂项相消法）
=1/2-(n+2)/[2^(n+2)(n+1)]