(1)解:由x
2+x-6=0,可得x=2或-3,
∵α、β是方程f(x)=0的兩個(gè)根(α>β),∴α=2,β=-3;
(2)解:∵g(x)=2x+1,∴a
n+1=g(a
n)=2a
n+1
∴a
n+1+1=2(a
n+1)
∵a
1=1,
∴{a
n+1}是以2為首項(xiàng),2為公比的等比數(shù)列
∴a
n+1=2
n,即a
n=2
n-1;
(3)證明:
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=
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∴a
n+1+3=

+3=
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,a
n+1-2=
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∴
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=ln
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=2ln
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=2b
n-1∴{b
n)是首項(xiàng)為ln
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=ln6,公比為2的等比數(shù)列
∴{b
n}的前n項(xiàng)和S
n=
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=(2
n-1)ln6.
分析:(1)先求出方程的根,再利用α、β是方程f(x)=0的兩個(gè)根(α>β),即可得到結(jié)論;
(2)證明{a
n+1}是以2為首項(xiàng),2為公比的等比數(shù)列,即可求得a
n;
(3)確定數(shù)列相鄰項(xiàng)的關(guān)系,可得等比數(shù)列,再利用等比數(shù)列的求和公式,即可得到結(jié)論.
點(diǎn)評(píng):本題考查數(shù)列與函數(shù)的關(guān)系,考查等比數(shù)列的判定,考查等比數(shù)列的通項(xiàng)與求和,屬于中檔題.