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已知数列an的前n项和为Sn,an=sn*s(n-1),n>=2,a1=2/9,a10=

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1 | 2010-1-8 18:00:27 | 显示全部楼层 |阅读模式

an=sn*s(n-1)an=(an+s(n-1))*s(n-1)an(1-s(n-1)) =s(n-1)2an=s(n-1)2/(1-s(n-1))=-s（n-1）-1+1/(1-s(n-1))=-(sn-an)-1+1/(1-s(n-1))sn+1=1/(1-s(n-1))sn+1-sn*s(n-1)-s(n-1)=1sn=sn*s(n-1)+s(n-1) 两边÷[sn*s(n-1)]1/(s(n-1))=1+1/sn所以1/s9=1+1/s101/s8=1+1/s91/s7=1+1/s8...1/s1=1+1/s2累加1/s1+...1/s9=9+1/s2+...+1/s101/s1=9+1/s10s1=a1=2/99/2=9+1/s10s10=-2/9s9=2/9a10=s10-s9=-4/9

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