已知数列an为等比数列,且首相a1=1/2,公比为1/2若bn=-log2an/an,求数列bn的前n项和sn
已知数列an为等比数列,且首相a1=1/2,公比为1/2若bn=-log2an/an,求数列bn的前n项和sn
数学人气:262 ℃时间:2019-08-22 08:14:13
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an=a1*(1/2)^(n-1)=(1/2)^n=2^(-n)-log2 an=-log2((1/2)^n )=n所以bn=n / (1/2)^n 则Sn=b1+b2+b3+.+bn-1+bn...(1)1/2Sn=(b1+b2+b3+.+bn-1+bn)*1/2...(2)两式想减,((1)中b2项-(2)中b1项.(1)中bn项-...
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