题目内容
15.已知数列{an}满足:a1=1,an+1=2an+1(n∈N*)(1)求数列{an}的通项公式;
(2)若数列{bn}满足4b1-1•4b2-1•4b3-1…4bn-1=(an+1)bn,证明:{bn}是等差数列.
分析 (1)通过对an+1=2an+1变形可得an+1+1=2(an+1),进而可得an+1=2n,从而可得结论;
(2)通过同底指数幂的运算性质,可得${4}^{{(b}_{1}+{b}_{2}+{…+b}_{n})-n}$=$({2}^{n})^{{b}_{n}}$,两边取对数得2[b1+b2+…+bn-n]=nbn,进而2[b1+b2+…+bn+1-(n+1)]=(n+1)bn+1,两式相减并整理得:(n-1)bn+1-nbn+2=0,进而nbn+2-(n+1)bn+1+2=0,再两式相减即得结论.
解答 (1)解:∵an+1=2an+1,
∴an+1+1=2(an+1),
∴数列{an+1}是公比为2的等比数列,
又∵a1=1,∴1+a1=2,
∴an+1=2n,
∴数列{an}的通项公式an=2n-1;
(2)证明:∵4b1-1•4b2-1•4b3-1…4bn-1=(an+1)bn,
∴${4}^{{(b}_{1}+{b}_{2}+{…+b}_{n})-n}$=$({2}^{n})^{{b}_{n}}$,
两边取对数,得:log2${4}^{{(b}_{1}+{b}_{2}+{…+b}_{n})-n}$=log2$({2}^{n})^{{b}_{n}}$,
∴2(b1+b2+…+bn)-2n=nbn,
即2[b1+b2+…+bn-n]=nbn,
2[b1+b2+…+bn+1-(n+1)]=(n+1)bn+1,
两式相减得:bn+1-1=$\frac{n+1}{2}$bn+1-$\frac{n}{2}$bn,
整理得:(n-1)bn+1-nbn+2=0,
∴nbn+2-(n+1)bn+1+2=0,
两式相减得:nbn+2-2nbn+1+nbn=0,
∴bn+2-2bn+1+bn=0,
即bn+2+bn=2bn+1,
∴数列{bn}是等差数列.
点评 本题考查利用数列的递推公式构造等比数列求解通项公式,利用数列的递推公式转化数列的和与项之间的关系,裂项求解数列的和的应用,注意解题方法的积累,属于中档题.
A. | 18 | B. | $\frac{1}{4}$ | C. | 16 | D. | $\frac{65}{4}$ |