根據(jù)
a = 2an + 3n -4
得到
an = 2a + 3(n-1) - 4 = 2a + 3n - 7
根據(jù)
a = 2an + 3n -4
得到
a - an + 3 = an + 3n -1
以及
an - a + 3 = a + 3(n-1) - 1 = a + 3n -4
將 an = 2a + 3n - 7 代入到 a - an + 3 = an + 3n -1 等式右端
a - an + 3 = 2a + 3n - 7 + 3n -1 = 2a + 6n - 8
至此
a - an + 3 = 2a + 6n - 8 = 2(a + 3n -4)
an - a + 3 = a + 3n -4
兩式做比值運算
[a - an + 3 ]/an - a + 3 = 2
因此 a(n+1)-an+3是等比數(shù)列 公比為 q = 2
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a1=-1,a(n+1)=2an+3n-4
a2 = 2a1 + 3 - 4 = -2 + 3 - 4 = -3
a2 - a1 + 3 = -3 + 1 + 3 = 1 (此為等比數(shù)列的首項)
根據(jù) a(n+1)-an+3是等比數(shù)列
a - an + 3 = [a2 - a1 + 3]*q^(n-1) = 2^(n-1)
a2 - a1 + 3 = 2^0
a3 - a2 + 3 = 2^1
a4 - a3 + 3 = 2^2
……
an - a + 3 = 2^(n-2)
以上共 n-1 個等式相加
an - a1 + 3(n-1) = 2^0 + 2^1 + …… + 2^(n-2)
an + 1 + 3n - 3 = 2^(n-1) -1
an = 2^(n-1) -3n + 1
根據(jù) an = 2^(n-1) - 3n + 1 ,則
n = 1 2 3 4 時 an < 0
n ≥ 5 時 an > 0
n ≤ 4 時
|a1|+|a2|+.+|an|
= -(a1 + a2 + …… + an)
= -[2^0 + 2^1 + …… +2^(n-1)] + 3(1 + 2 …… + n) - 1*n
= -2^n + 1 + 3n(1+n)/2 - n
= n(3n+1)/2 - 2^n + 1
n = 4 時
S4 = 4(3*4+1)/2 - 2^4 + 1 = 11
n ≥ 5 時
|a1|+|a2|+.+|an|
=S4 + a5 + a6 + …… + a7
= 11 + [2^4 + 2^5 + …… + 2^(n-1)] - 3 (5 + 6 + …… + n) + (n-4)
= 11 + 2^4 * [2^(n-4) -1] - 3(5 + n)(n-4)/2 + (n-4)
= 11 + 2^n - 16 - 3(n-4)(n+5)/2 + (n-4)
= 2^n - 5 - (n-4)(3n+13)/2