题目内容
11.已知等差数列{an}的公差d=$\int_{-\frac{π}{2}}^{\frac{π}{2}}{cosxdx}$,a42-a22=56;等比数列{bn}满足:b1=1,b2b4b6=512,n∈N*.(1)求数列{an}和{bn}的通项公式;
(2)设{an}的前n项和为Sn,令cn=$\left\{{\begin{array}{l}{\frac{2}{S_n},n为奇数}\\{{b_n},n为偶数}\end{array}}$,求c1+c2+c3+…+c2n.
分析 (1)通过定积分的计算可得$d=\int_{-\frac{π}{2}}^{\frac{π}{2}}{cosxdx}=2$,通过$a_4^2-a_2^2=({a_4}+{a_2})({a_4}-{a_2})=2{a_3}•2d=56$可得a3=7,进而可得a1=3,即得an=2n+1;设等比数列{bn}的公比为q,通过等比中项的性质及${b_2}{b_4}{b_6}=b_4^3=512$,可得b4=8,进而可得q=2,即得${b_n}={b_1}{q^{n-1}}={2^{n-1}}$;
(2)通过a1=3、an=2n+1得${c_n}=\left\{{\begin{array}{l}{\frac{1}{n}-\frac{1}{n+2},n为奇数}\\{{2^{n-1}},n为偶数}\end{array}}\right.$,利用并项相加法计算即可.
解答 解:(1)公差$d=\int_{-\frac{π}{2}}^{\frac{π}{2}}{cosxdx}=2$,
∵$a_4^2-a_2^2=({a_4}+{a_2})({a_4}-{a_2})=2{a_3}•2d=56$,
∴a3=7,
∴a1+2d=7,∴a1=3,
∴an=3+2(n-1)=2n+1;
设等比数列{bn}的公比为q,
∵${b_2}{b_4}{b_6}=b_4^3=512$,
∴b4=8,即b1q3=8,∴q=2,
即${b_n}={b_1}{q^{n-1}}={2^{n-1}}$;
(2)由a1=3,an=2n+1得:Sn=n(n+2),
∴${c_n}=\left\{{\begin{array}{l}{\frac{2}{n(n+2)},n为奇数}\\{{2^{n-1}},n为偶数}\end{array}}\right.$,
即${c_n}=\left\{{\begin{array}{l}{\frac{1}{n}-\frac{1}{n+2},n为奇数}\\{{2^{n-1}},n为偶数}\end{array}}\right.$,
∴c1+c2+c3+…c2n=(c1+c3+…c2n-1)+(c2+c4+…c2n)
=$[(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+…+(\frac{1}{2n-1}-\frac{1}{2n+1})]+(2+{2^3}+…{2^{2n-1}})$
=$1-\frac{1}{2n+1}+\frac{{2(1-{4^n})}}{1-4}=\frac{2n}{2n+1}+\frac{2}{3}({4^n}-1)$.
点评 本题考查求数列的通项及前n项和,涉及到定积分的计算等知识,考查分类讨论的思想,利用并项相加法是解决本题的关键,注意解题方法的积累,属于中档题.
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