题目内容
19.已知数列{an}满足${S_n}=2{n^2}+n-1$,则通项an=$\left\{\begin{array}{l}2,n=1\\ 4n-1,n≥2\end{array}\right.$.分析 利用“当n=1时,a1=S1.当n≥2时,an=Sn-Sn-1”即可得出.
解答 解:当n=1时,a1=S1=2+1-1=2.
当n≥2时,an=Sn-Sn-1=2n2+n-1-[2(n-1)2+(n-1)-1]=4n-1.
∴an=$\left\{\begin{array}{l}2,n=1\\ 4n-1,n≥2\end{array}\right.$.
故答案为:$\left\{\begin{array}{l}2,n=1\\ 4n-1,n≥2\end{array}\right.$.
点评 本题考查了利用“当n=1时,a1=S1.当n≥2时,an=Sn-Sn-1”求数列通项公式,属于基础题.
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