已知数列{an}满足an=2an-1+2n-1(n≥2),a1=5,bn=an−12n (Ⅰ)证明:{bn}为等差数列; (Ⅱ)求数列{an}的前n项和Sn.
已知数列{an}满足an=2an-1+2n-1(n≥2),a1=5,bn=
(Ⅰ)证明:{bn}为等差数列;
(Ⅱ)求数列{an}的前n项和Sn.
an−1 |
2n |
(Ⅰ)证明:{bn}为等差数列;
(Ⅱ)求数列{an}的前n项和Sn.
数学人气:392 ℃时间:2019-08-19 08:23:03
优质解答
(I)证明:∵an=2an-1+2n-1(n≥2),∴an−1=2(an−1−1)+2n,∴an−12n=an−1−12n−1+1.∴bn=bn-1+1.∴{bn}是首项为a1−12=5−12=2,公差为1的等差数列;(II)由(I)可得bn=2+(n-1)×1=n+1,∴an−12n=...
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