an+1+an=c(c为关于n的指数函数)的递推式,怎么求an通项

Question

an+1+an=c(c为关于n的指数函数)的递推式,怎么求an通项

Answer 1

你这个问题,即
a(n+1) = -a(n) +c *q^n.
(c,q为常数).
更一般的递推公式为
a(n+1) = k a(n) +c *q^n.
(k,c,q为常数).
= = = = = = = = =
1.a(n+1) =k a(n) +c *q^n ,
(k,c,q为常数,且 k 不等于1,k 不等于q ).
令 a(n+1) +x *q^(n+1) =k [ a(n) +x *q^n],
则 a(n+1) =k a(n) +(k-q) x *q^n.
比较系数,得
(k-q) x =c.
解得 x =c/(k-q).
所以 a(n+1) +x *q^(n+1) =k [ a(n) +x *q^n ].
所以 { a(n) +x *q^n }是等比数列,且
首项为 a(1) +x *q^1 = a(1) +x.
公比为 k.
所以 a(n) +x *q^n =[ a(1) +x ] *k^(n-1).
即 a(n) = [ a(1) +x ] *k^(n-1) -x *q^n,
其中 x= c/(k-q).
= = = = = = = = = =
k不等于1,且k不等于q.用待定系数法.
= = = = = = = = = =
2.a(n+1) =k a(n) +c
(k,c为常数,且 k 不等于1).
令 a(n+1) +x =k [a(n) +x],
则 a(n+1) =k a(n) +(k-1)x.
.
= = = = = = = = =
q=1,还是待定系数法.
3.a(n+1)=q a(n) +c *q^n.
(q,c为常数,且q不等于1).
(1) a(2) =q a(1) +cq,
a(3) =q a(2) +c *q^2
=q^2 *a(1) +2c *q^2.
(2) 设 当n=m>=3时,有
a(m) =q^(m-1) *a(1) +(m-1)c *q^(m-1).
则 当 n=m+1时,
a(m+1) =q a(m) +c *q^m
=q [q^(m-1) *a(1) +(m-1)c *q^(m-1)]
=q^m *a(1) +mc *q^m.
综上,由数学归纳法知,
a(n) =q^n *a(1) +nc *q^n.n>=2.
= = = = = = = =
按书上的数学归纳法那样做,格式我忘了.
这题中 k=q.
题1用数学归纳法很难做,这题用待定系数法做不出.
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4.a(n+1) =a(n) +c *q^n.
(c,q为常数,且q不等于1).
因为 a(n+1) -a(n) =c *q^n,
所以 a(n) -a(n-1) =c *q^(n-1),
...
a(2) -a(1) =c *q.
逐项相加,得
a(n+1) -a(1) =cq *(1 -q^n) /(1-q).
即 a(n+1) =a(1) +cq/(1-q) *(1 -q^n) .
即 a(n) =a(1) +cq/(1-q) *[ 1 -q^(n-1) ],n>=2 .
= = = = = = = = = =
k=1 是好事情.
5.a(n+1)=a(n)+c.
= = = = = = = = =
等差数列.