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数列Cn=(n+2)/[n(n+1)]2^n的Sn
人气:109 ℃ 时间:2020-05-24 04:11:25
解答
裂项相消:Cn=[(n+2)/n-(n+2)/(n+1)]2^n=2/(n2^n)-1/((n+1)2^n)=1/(n2^(n-1)-1/((n+1)2^n),因此Sn=1-1/(2*2)+1/(2*2)-1/(3*2^2)+1/(3*2^2)-1/(4*2^3)+...+1/(n2^(n-1)-1/((n+1)2^n)=1-1/((n+1)2^n)...
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