题目内容
根据如图所示的程序框图,将输出的值依
次分别记为;,…,,….
(Ⅰ)分别求数列和的通项公式;
(Ⅱ)令,求数列的前项和,
其中.
;
(Ⅰ)由框图,知数列
∴
由框图,知数列中, ∴
∴ ∴数列是以3为首项,3为公比的等比数列
∴ ∴
(Ⅱ)=
=1×(3-1)+3×(32-1)+…+(2k-1)(3k-1)
=1×3+3×32+…+(2k-1)·3k-[1+3+…+(2k-1)]
记 1×3+3×32+…+(2k-1)·3k,①
则1×32+3×33+…+(2k-1)×3k+1 ②
①-②,得-2Sk=3+2·32+2·33+…+2·3k-(2k-1)·3k+1
=2(3+32+…+3k)-3-(2k-1)·3k+1
=2×
=
∴
又1+3+…+(2k-1)=k2
∴
∴
由框图,知数列中, ∴
∴ ∴数列是以3为首项,3为公比的等比数列
∴ ∴
(Ⅱ)=
=1×(3-1)+3×(32-1)+…+(2k-1)(3k-1)
=1×3+3×32+…+(2k-1)·3k-[1+3+…+(2k-1)]
记 1×3+3×32+…+(2k-1)·3k,①
则1×32+3×33+…+(2k-1)×3k+1 ②
①-②,得-2Sk=3+2·32+2·33+…+2·3k-(2k-1)·3k+1
=2(3+32+…+3k)-3-(2k-1)·3k+1
=2×
=
∴
又1+3+…+(2k-1)=k2
∴
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