题目内容
13.已知公比q不为1的等比数列{an}的首项${a_1}=\frac{1}{2}$,前n项和为Sn,且a4+S4,a5+S5,a6+S6成等差数列.(1)求数列{an}的通项公式;
(2)对n∈N+,在an与an+1之间插入n个数,使这n+2个数成等差数列,记插入的这n个数的和为{bn},求数列{bn}的前n项和Tn.
分析 (I)通过2(a5+S5)=a4+S4+a6+S6化简得2q2-3q+1=0,进而计算可得结论;
(II)通过bn=$\frac{3}{4}$n•$\frac{1}{{2}^{n}}$,写出Tn、$\frac{1}{2}•$Tn的表达式,利用错位相减法计算即得结论.
解答 解:(I)由题可知:2(a5+S5)=a4+S4+a6+S6,
化简得:2a6-3a5+a4=0,
∴2q2-3q+1=0,
解得:q=$\frac{1}{2}$或q=1(舍),
∴an=$\frac{1}{2}•$$\frac{1}{{2}^{n-1}}$=$\frac{1}{{2}^{n}}$;
(II)依题意bn=$\frac{n({a}_{n}+{a}_{n+1})}{2}$=$\frac{3}{4}$n•$\frac{1}{{2}^{n}}$,
∴${T_n}=\frac{3}{4}[1×\frac{1}{2}+2×{(\frac{1}{2})^2}+3×{(\frac{1}{2})^3}+…+(n-1){(\frac{1}{2})^{n-1}}+n×{(\frac{1}{2})^n}]$,
$\frac{1}{2}{T_n}=\frac{3}{4}[1×{(\frac{1}{2})^2}+2×{(\frac{1}{2})^3}+3×{(\frac{1}{2})^4}+…+(n-1){(\frac{1}{2})^n}+n×{(\frac{1}{2})^{n+1}}]$,
两式相减得:$\frac{1}{2}{T_n}=\frac{3}{4}[\frac{1}{2}+{(\frac{1}{2})^2}+{(\frac{1}{2})^3}+…+{(\frac{1}{2})^n}-n{(\frac{1}{2})^{n+1}}]$
=$\frac{3}{4}$•($\frac{1}{2}$•$\frac{1-\frac{1}{{2}^{n}}}{1-\frac{1}{2}}$-n•$\frac{1}{{2}^{n+1}}$),
∴Tn=$\frac{3}{2}$(1-$\frac{1}{{2}^{n}}$-$\frac{n}{{2}^{n+1}}$).
点评 本题考查数列的通项及前n项和,考查运算求解能力,注意解题方法的积累,属于中档题.
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