已知数列｛an｝的前n项和为Sn,且S(n+1)=4an+2,a1=1,(1)设bn=a(n+1)-2an,求证：｛bn

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已知数列｛an｝的前n项和为Sn,且S(n+1)=4an+2,a1=1,(1)设bn=a(n+1)-2an,求证：｛bn｝为等比数列； (2)
已知数列｛an｝的前n项和为Sn,且S(n+1)=4an+2,a1=1,
(1)设bn=a(n+1)-2an,求证：｛bn｝为等比数列；
(2)设Cn=an/2^n,求证：｛Cn｝为等差数列.

1.S(n+1) = 4an + 2 .(1)
则：Sn = 4a(n-1) + 2 .(2)
两式相减：a(n+1) = 4an - 4a(n-1)
a(n+1) - 2an = 2[an - 2a(n-1)]
∴{a(n+1) - 2an}={bn} 是等比数列,且公比q=2
2.∵S2=a1 + a2 = 4a1 + 2
∴a2=5
则：b1=a2 - 2a1 =3
bn=3×2^(n-1) ,n∈N
即：a(n+1) - 2an =3×2^(n-1)
则：a(n+1)=3×2^(n-1) + 2an
C(n+1) - Cn = a(n+1)/2^(n+1) - an/2^n =[3×2^(n-1) + 2an]/[2^(n+1)] - (2an)/2^n
=[3×2^(n-1)]/[2^(n+1)]
= 3/4
∴｛Cn｝为等差数列,且公差d=3/4

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