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:
1
S1
+
1
S2
+…+
1
Sn
n
n+1
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=

n(1+2n-1)
2
=n2

(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
n+1
=1-
1
n+1
=
n
n+1

1
S1
+
1
S2
+…+
1
Sn
n
n+1

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