{an}中,a1=1,a2=3,a(n+2)=4a(n+1)-4an 求证an/2^(n-1)为等差数列 n+2和n+1为下标

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{an}中,a1=1,a2=3,a(n+2)=4a(n+1)-4an求证an/2^(n-1)为等差数列n+2和n+1为下标{an}中,a1=1,a2=3,a(n+2)=4a(n+1)-4an求证an/

{an}中,a1=1,a2=3,a(n+2)=4a(n+1)-4an 求证an/2^(n-1)为等差数列 n+2和n+1为下标
{an}中,a1=1,a2=3,a(n+2)=4a(n+1)-4an 求证an/2^(n-1)为等差数列 n+2和n+1为下标

{an}中,a1=1,a2=3,a(n+2)=4a(n+1)-4an 求证an/2^(n-1)为等差数列 n+2和n+1为下标
由a(n+2)=4a(n+1)-4an得到
a(n+2)-2a(n+1)=2(a(n+1)-2an)
因此得到数列 {a(n+1)-2an}是首项为1,等比为2的数列
故有a(n+1)-2an=2^(n-1)
因此上式左右同时除以2^(n)
得到a(n+1)/2^n-an/2^(n-1)=1/2
故数列an/2^(n-1)是首项为1,等差为1/2的等差数列.

∵a(n+2)=4a(n+1)-4an
∴a(n+2)-2a(n+1)
=2a(n+1)-4an
=2[a(n+1)-2an]
∴[a(n+2)-2a(n+1)]/[a(n+1)-2an]=2
∴{a(n+1)-an}为等比数列,公比为2
又a1=1,a2=3,a2-2a1=1
∴a(n+1)-2an
=(a2-2a1)*2^(...

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∵a(n+2)=4a(n+1)-4an
∴a(n+2)-2a(n+1)
=2a(n+1)-4an
=2[a(n+1)-2an]
∴[a(n+2)-2a(n+1)]/[a(n+1)-2an]=2
∴{a(n+1)-an}为等比数列,公比为2
又a1=1,a2=3,a2-2a1=1
∴a(n+1)-2an
=(a2-2a1)*2^(n-1)
=2^(n-1)

∴a(n+1)/2^n-an/2^(n-1)
=[a(n+1)-2an]/2^n
=2^(n-1)/2^n
=1/2
∴{an/2^(n-1)}为等差数列,公差为2
参考http://58.130.5.100//

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