题目内容
11.已知数列{an}的前n项和为Sn和为Sn,S1=-$\frac{1}{4}$,an-4SnSn-1=0(n≥2)(1)若bn=$\frac{1}{{S}_{n}}$,证明{bn}是等差数列;
(2)求数列{an}的通项公式.
分析 (1)根据an=Sn-Sn-1,结合n≥2时,an-4SnSn-1=0,可得Sn-Sn-1=4SnSn-1,两边同除SnSn-1可得结论;
(2)根据(1)可得Sn=$-\frac{1}{4n}$,结合S1=-$\frac{1}{4}$,n≥2时,an-4SnSn-1=0,可得数列{an}的通项公式.
解答 证明:(1)∵n≥2时,an-4SnSn-1=0,
∴Sn-Sn-1-4SnSn-1=0
∴Sn-Sn-1=4SnSn-1,
∴$\frac{1}{{S}_{n}}$-$\frac{1}{{S}_{n-1}}$=-4,
又∵S1=-$\frac{1}{4}$,
∴$\frac{1}{{S}_{1}}$=-4,
又∵bn=$\frac{1}{{S}_{n}}$,
∴数列{bn}是以-4为首项,-4为公差的等差数列;
解:(2)由(1)知bn=$\frac{1}{{S}_{n}}$=-4n,
∴Sn=$-\frac{1}{4n}$;
∵n≥2时,an-4SnSn-1=0,
∴an=4SnSn-1=$\frac{1}{4n(n-1)}$,
当n=1时,$\frac{1}{4n(n-1)}$无意义,
∴an=$\left\{\begin{array}{l}-\frac{1}{4},n=1\\ \frac{1}{4n(n-1)},n≥2\end{array}\right.$.
点评 本题考查等差数列的证明,考查数列的求和与通项,正确运用数列递推式是关键.
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