已知各項都不相等的等差數(shù)列{a
n}的前6項和為60,且a
6為a
1和a
21的等比中項.
(1)求數(shù)列{a
n}的通項公式.
(2)若數(shù)列{b
n}滿足b
n+1-b
n=a
n(n∈N
*),且b
1=3,求數(shù)列{
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}的前n項和T
n.
(1) a
n=2n+ (2) T
n=
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(1)設(shè)等差數(shù)列{a
n}的公差為d(d≠0),
則
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解得
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∴a
n=2n+3.
(2)由b
n+1-b
n=a
n,
∴b
n-b
n-1=a
n-1(n≥2,n∈N
*),
b
n=(b
n-b
n-1)+(b
n-1-b
n-2)+…+(b
2-b
1)+b
1=a
n-1+a
n-2+…+a
1+b
1=n(n+2),
當(dāng)n=1時,b
1=3也適合上式,
∴b
n=n(n+2)(n∈N
*).
∴
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=
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=
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(
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-
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),
T
n=
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(1-
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+
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-
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+…+

-
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)
=
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(
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-
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-
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)=
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.
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