题目内容
7.已知等差数列{an},满足a2=2,a4=4.(1)求数列{an}的通项公式;
(2)求数列$\left\{{\frac{1}{{{a_n}{a_{n+2}}}}}\right\}$的前n项和.
分析 (1)利用等差数列的通项公式即可得出;
(2)利用“裂项求和”即可得出.
解答 解:(1)设等差数列{an}的公差为d,
则$\left\{\begin{array}{l}{a_2}={a_1}+d=2\\{a_4}={a_1}+3d=4\end{array}\right.$,解得$\left\{{\begin{array}{l}{{a_1}=1}\\{d=1}\end{array}}\right.$,
∴an=a1+(n-1)d=n,
故an=n.
(2)$\frac{1}{{{a_n}{a_{n+2}}}}=\frac{1}{n(n+2)}=\frac{1}{2}(\frac{1}{n}-\frac{1}{n+2})$,
故$\frac{1}{{{a_1}{a_3}}}+\frac{1}{{{a_2}{a_4}}}+…+\frac{1}{{{a_n}{a_{n+2}}}}$
=$\frac{1}{2}[(1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+…+(\frac{1}{n-1}-\frac{1}{n+1})+(\frac{1}{n}-\frac{1}{n+2})]$
=$\frac{1}{2}(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2})=\frac{1}{2}×(\frac{3}{2}-\frac{1}{n+1}-\frac{1}{n+2})=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}$.
点评 本题考查了等差数列的通项公式、“裂项求和”方法,考查了推理能力与计算能力,属于中档题.
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