待定系數(shù)法：
首先求得a3=27,a2=9,a1=2
設(shè)一整數(shù)λ.我們可以先構(gòu)造一個{(an+λ)/(2^n)}數(shù)列.
(an+λ)/(2^n) - (a(n-1)+λ)/(2^(n-1)) = (an+λ-2a(n-1)-2λ)/(2^n)
將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)
故此時可令λ=1.
從而 (an+1)/(2^n) - (a(n-1)+1)/(2^(n-1) = 1
即數(shù)列{(an+λ)/(2^n)}是以1為公差,(a1+1)/2=3/2為首項的等差數(shù)列.
其通項為：(an+λ)/(2^n)=2/3+n-1=1/2+n 移項后可得：
an=(1/2+n)2^n -1
代入a1,a2,a3進行驗算,亦符合本通式.
建議你在紙上寫一寫,可能這個看著并不是很清楚.