(1)證明:∵a
1=1,3tS
n-(2t+3)S
n-1=3t(n≥2,n∈N
*)①
∴3tS
n-1-(2t+3)S
n-2=3t(n≥3,n∈N*)②
①②兩式相減得
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又n=2時,
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∴a
n是以1為首項,
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為公比的等比數列.
(2)解:∵
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,∴
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,∴
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∴b
n是以1為首項,
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為公差的等差數列,∴
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∴b
1b
2-b
2b
3+b
3b
4-b
4b
5+…+b
2n-1b
2n-b
2nb
2n+1(n∈N*)
=b
2(b
1-b
3)+b
4(b
3-b
4)+…+b
2n(b
2n-1-b
2n+1)
=
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.
分析:(1)因為a
n=S
n-S
n-1(n≥2,n∈N
*),所以在3tS
n-(2t+3)S
n-1=3t的基礎上,用n-1替換n構造與它類似的關系式;然后利用作差法求出a
n與a
n-1的關系式,進而可整理為等比數列形式;但不要忘掉未含項的檢驗.
(2)由(1)知{a
n}的公比f(t),又b
n=f(
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),則可找到b
n與b
n-1的關系,進而可整理為等差數列形式;則由等差數列通項公式可求b
n;代數式b
1b
2-b
2b
3+b
3b
4-b
4b
5+…+b
2n-1b
2n-b
2nb
2n+1的求值,可利用分組的方法,把它轉化到等差數列的性質與前n項和公式上去,則問題解決.
點評:若數列{a
n}的前n項和為S
n,則a
n=S
n-S
n-1(n≥2,n∈N
*)是實現前n項和S
n向通項a
n轉化的橋梁與紐帶,進而可結合等差數列、等比數列的定義與性質解決問題.