已知数列an中 a1=1
作者&投稿：乐正韵 （若有异议请与网页底部的电邮联系）
已知数列an中a1=1且an+1=an/an+1~
2^(n+1)a(n+1) = (2^n)an + 2n+3
2^(n+1)a(n+1) -(2^n)an = 2n+3
(2^n)an -(2^(n-1))a(n-1) = 2n+1
(2^n)an -(2^1)a1 = 5+7+....+2n+1
2^n. an = (n+3)(n-1) +1
= n^2+2n-2
an =[ n(n+1) +n-2 ].2^n
= n(n+1).2^n + n.2^n - 2.2^n
Sn =a1+a2+...+an
= S + S' - 4(2^n-1)
S' = 1.2^1+ 2.2^2+....+n.2^n (1)
2S' = 1.2^2+ 2.2^3+....+n.2^(n+1) (2)
(2)-(1)
S' = n.2^(n+1) - (2+2^2+...+2^n)
=n.2^(n+1) - 2(2^n-1)
= 2+ (2n-2).2^n
S = (1.2).2^1 + (2.3).2^2+......+n(n+1).2^n (3)
2S = (1.2).2^2 + (2.3).2^3+......+n(n+1).2^(n+1) (4)
(2)-(1)
S = n(n+1).2^(n+1) -2[ 1.2^1 + 2.2^2+3.2^3+....+n.2^n ]
= n(n+1).2^(n+1) -2S'
Sn =a1+a2+...+an
= S + S' - 4(2^n-1)
=n(n+1).2^(n+1) - S' - 4(2^n-1)
=n(n+1).2^(n+1) - [2+ (2n-2).2^n] - 4(2^n-1)
= 2 + (n^2-1)2^(n+1)
令an = bn / (2 ^ n)
代入递推公式，化简得b(n + 1) - bn = 2n + 3
b1 = 2 * a1 = 1
可得bn = n ^ 2 + 2n - 2
即an = (n ^ 2 + 2n - 2) / (2 ^ n)
再令cn = -(n ^ 2 + 6n + 8) / (2 ^ n)
那么an = cn - c(n - 1)
即Sn = cn - c0 = 8 - (n ^ 2 + 6n + 8) / (2 ^ n)
cn分子中的各系数可通过待定系数法求得
由题，1/a(n+1)=[a(n)+1]/a(n)
即，1/a(n+1)=1+1/a(n)
所以，1/a(n+1)－1/a(n)＝1
又a(1)＝1，1/a(1)＝1
所以，数列{1/an}为等差数列，首项＝1，公差＝1
所以，1/a(n)＝1＋(n-1)×1＝n
即，an＝1/n
所以，数列{an}的通项公式为 an＝1/n
a(n+1)=2an/(an + 2)
1/a(n+1)= (an + 2)/2an = 1/2 + 1/an
1/a(n+1) - 1/an = 1/2 所以1/an 为等差数列，公差是1/2
1/an = 1/1 + 1*(n-1)/2 = (n+1)/2
an = 2/(n+1)
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2^(n+1)a(n+1) = (2^n)an + 2n+3
2^(n+1)a(n+1) -(2^n)an = 2n+3
(2^n)an -(2^(n-1))a(n-1) = 2n+1
(2^n)an -(2^1)a1 = 5+7+....+2n+1
2^n. an = (n+3)(n-1) +1
= n^2+2n-2
an =[ n(n+1) +n-2 ].2^n
= n(n+1).2^n + n.2^n - 2.2^n
Sn =a1+a2+...+an
= S + S' - 4(2^n-1)
S' = 1.2^1+ 2.2^2+....+n.2^n (1)
2S' = 1.2^2+ 2.2^3+....+n.2^(n+1) (2)
(2)-(1)
S' = n.2^(n+1) - (2+2^2+...+2^n)
=n.2^(n+1) - 2(2^n-1)
= 2+ (2n-2).2^n
S = (1.2).2^1 + (2.3).2^2+......+n(n+1).2^n (3)
2S = (1.2).2^2 + (2.3).2^3+......+n(n+1).2^(n+1) (4)
(2)-(1)
S = n(n+1).2^(n+1) -2[ 1.2^1 + 2.2^2+3.2^3+....+n.2^n ]
= n(n+1).2^(n+1) -2S'
Sn =a1+a2+...+an
= S + S' - 4(2^n-1)
=n(n+1).2^(n+1) - S' - 4(2^n-1)
=n(n+1).2^(n+1) - [2+ (2n-2).2^n] - 4(2^n-1)
= 2 + (n^2-1)2^(n+1)
令an = bn / (2 ^ n)
代入递推公式，化简得b(n + 1) - bn = 2n + 3
b1 = 2 * a1 = 1
可得bn = n ^ 2 + 2n - 2
即an = (n ^ 2 + 2n - 2) / (2 ^ n)
再令cn = -(n ^ 2 + 6n + 8) / (2 ^ n)
那么an = cn - c(n - 1)
即Sn = cn - c0 = 8 - (n ^ 2 + 6n + 8) / (2 ^ n)
cn分子中的各系数可通过待定系数法求得
