已知数列{an}满足a1=1,a2=20,a(n+2)=2a(n+1)+8an,求an通项和前n项和Sn,补充：第一小题已证明{a(n+1)+2an}是等比数列 且a(n+2)中n+2为下标

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已知数列{an}满足a1=1,a2=20,a(n+2)=2a(n+1)+8an,求an通项和前n项和Sn,补充：第一小题已证明{a(n+1)+2an}是等比数列且a(n+2)中n+2为下标已知数列{a

已知数列{an}满足a1=1,a2=20,a(n+2)=2a(n+1)+8an,求an通项和前n项和Sn,补充：第一小题已证明{a(n+1)+2an}是等比数列 且a(n+2)中n+2为下标
已知数列{an}满足a1=1,a2=20,a(n+2)=2a(n+1)+8an,求an通项和前n项和Sn,补充：第一小题已证明{a(n+1)+2an}是等比数列 且a(n+2)中n+2为下标

已知数列{an}满足a1=1,a2=20,a(n+2)=2a(n+1)+8an,求an通项和前n项和Sn,补充：第一小题已证明{a(n+1)+2an}是等比数列 且a(n+2)中n+2为下标
a(n+2)=2a(n+1)+8a(n),
r^2 = 2r+8,0=r^2 - 2r - 8 = (r-4)(r+2),r=4 或r=-2.
a(n+2) + 2a(n+1) = 4a(n+1) + 8a(n) = 4[a(n+1)+2a(n)],
{a(n+1)+2a(n)}是首项为a(2)+2a(1)=22,公比为4的等比数列.
a(n+1)+2a(n)=22*4^(n-1),.(*)
a(n+2) - 4a(n+1) = -2a(n+1) + 8a(n) = -2[a(n+1) - 4a(n)],
{a(n+1) - 4a(n)}是首项为a(2)-4a(1)=16,公比为（-2）的等比数列.
a(n+1)-4a(n) = 16*(-2)^(n-1),.(**)
式(*) - (**),有,
2a(n) + 4a(n) = 6a(n) = 22*4^(n-1) - 16*(-2)^(n-1),
a(n) = [11*4^(n-1) - 8*(-2)^(n-1)]/3,
s(n) = (11/3)[1+4+...+4^(n-1)] + (-8/3)[1+(-2) + ...+ (-2)^(n-1)]
=(11/3)[4^n - 1]/(4-1) -(8/3)[(-2)^n - 1]/(-2-1)
=(11/9)[4^n - 1] + (8/9) [ (-2)^n - 1]
=(11/9)4^n + (8/9)(-2)^n - 19/9

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