问题
解答题
已知数列{an},a1=3,an+1=4an-3 (Ⅰ)设bn=1og2(an-1),求数列{bn}的前n项和Sn (Ⅱ)在(Ⅰ)的条件下,求证:
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答案
(Ⅰ)∵an+1=4an-3,∴an+1-1=4(an-1)
∵a1=3,∴a1-1=2,
∴{an-1}是以2为首项,4为公比的等比数列
∴an-1=2×4n-1=22n-1,
∵bn=1og2(an-1),∴bn=2n-1,
∴数列{bn}是以1为首项,2为公差的等差数列
∴Sn=
=n2;n(1+2n-1) 2
(Ⅱ)证明:
+1 S1
+…+1 S2
=1 Sn
+1 12
+…+1 22
>1 n2
+1 1×2
+…+1 2×3 1 n(n+1)
=1-
+1 2
-1 2
+…+1 3
-1 n
=1-1 n+1
=1 n+1 n n+1
∴
+1 S1
+…+1 S2
>1 Sn
.n n+1