已知数列{log
2(a
n-1)}(n∈N
*)为等差数列,且a
1=3,a
3=9.
(Ⅰ)求数列{a
n}的通项公式;
(Ⅱ)证明
+
+…+
<1.
人气:480 ℃ 时间:2020-05-06 00:35:54
解答
(I)设等差数列{log
2(a
n-1)}的公差为d.
由a
1=3,a
3=9得2(log
22+d)=log
22+log
28,即d=1.
所以log
2(a
n-1)=1+(n-1)×1=n,即a
n=2
n+1.
(II)证明:因为
=
=
,
所以
+
+…+
=
+
+
+…+
=
=1-
<1,
即得证.
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