有点难的数列题设A1=1,A(n+1)=2An+n*2的n次方+(-1)的n次方,求通项An
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有点难的数列题设A1=1,A(n+1)=2An+n*2的n次方+(-1)的n次方,求通项An
有点难的数列题
设A1=1,A(n+1)=2An+n*2的n次方+(-1)的n次方,求通项An
有点难的数列题设A1=1,A(n+1)=2An+n*2的n次方+(-1)的n次方,求通项An
先给个答案.An=n(n-1)*[2^(n-2)]+[2^n-(-1)^n]/3.(n=1,2,3,4,...).原式两边同除以2^n,并设Bn=An/[2^(n-1)],则有B(n+1)=Bn+n+(-1/2)^n,求出B1,B2,B3,B4,B5,.累加得：Bn=[n(n-1)/2]+[2+(-1/2)^(n-1)]/3.还原即得通项.可求得：A1=1,A2=3,A3=15,A4=53,我求的通项可检验.
An=n*(n-1)*2的(n-2)次方+(1/3)*2的(n+2)次方-(1/3)*(-1)的n次方
你的题目表述不清楚，我按自己的理解给出
A(n+1) = 2A(n) + n2^n + (-1)^n = 2A(n) - (n+1)2^n + n2^n + 2^n + (-1)^n,
A(n+1) + (n+1)2^n = 2[A(n) + n2^(n-1)] + 2^n + (-1)^n,
b(n) = A(n) + n2^(n-1),
b(n+1) = 2b(n) + 2^n + (-1)^n,
b...
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A(n+1) = 2A(n) + n2^n + (-1)^n = 2A(n) - (n+1)2^n + n2^n + 2^n + (-1)^n,
A(n+1) + (n+1)2^n = 2[A(n) + n2^(n-1)] + 2^n + (-1)^n,
b(n) = A(n) + n2^(n-1),
b(n+1) = 2b(n) + 2^n + (-1)^n,
b(n+1)/2^n = b(n)/2^(n-1) + (-1/2)^n,
c(n) = b(n)/2^(n-1),
c(n+1) = c(n) + (-1/2)^n,
c(n) = c(n-1) + (-1/2)^(n-1),
...
c(2) = c(1) + (-1/2)^1,
c(n+1) = c(1) + (-1/2)^1 + (-1/2)^2 + ... + (-1/2)^n
= b(1)/1 + (-1/2)[1 + (-1/2)^1 + ... + (-1/2)^(n-1)]
= A(1)+1 + (-1/2)[1 - (-1/2)^n]/[1-(-1/2)]
= 1+1 + (1/2)[(-1/2)^n - 1]/(3/2)
= 2 + [(-1/2)^n - 1]/3
b(n) = 2^(n-1)c(n) = 2^(n-1){2 + [(-1/2)^(n-1) - 1]/3}
= 2^n + [(-1)^(n-1) - 2^(n-1)]/3,
A(n) = b(n) - n2^(n-1) = 2^n + [(-1)^(n-1) - 2^(n-1)]/3 - n2^(n-1)
= 2^(n-1)[2 - 1/3 - n] - (-1)^n/3
= [2^(n-1)(5-3n) - (-1)^n]/3, n = 1,2,...
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A1=1,A(n+1)=2An+n*2的n次方+(-1)的n次方
楼主：n*2的n次方+(-1)的n次方，打出来下，看得方便，OK?