Question

数列{an}中,an=2a(n-1)+2^n+1,a3=27 求an通项公式?

Answer 1

待定系数法：
首先求得a3=27,a2=9,a1=2
设一整数λ.我们可以先构造一个{(an+λ)/(2^n)}数列.
(an+λ)/(2^n) - (a(n-1)+λ)/(2^(n-1)) = (an+λ-2a(n-1)-2λ)/(2^n)
将an=2a(n-1)+2^n+1 代入上式,可得到
(an+λ)/(2^n) - (a(n-1)+λ)/(2^(n-1) = 1 + (1-λ)/(2^n)
故此时可令λ=1.
从而 (an+1)/(2^n) - (a(n-1)+1)/(2^(n-1) = 1
即数列{(an+λ)/(2^n)}是以1为公差,(a1+1)/2=3/2为首项的等差数列.
其通项为：(an+λ)/(2^n)=2/3+n-1=1/2+n 移项后可得：
an=(1/2+n)2^n -1
代入a1,a2,a3进行验算,亦符合本通式.
建议你在纸上写一写,可能这个看着并不是很清楚.