a1=2a(n+1)=λAn+λ^(N+1)+(2-λ)2^N.求解题思路.求通项共识AND求前n项和.要思路

Question

a1=2 a(n+1) =λAn + λ^(N+1)+(2-λ)2^N.求解题思路.
求通项共识AND求前n项和.要思路

Answer 1

a[n+1]=λa[n]+λ^(n+1)+(2-λ)2^n
两边除以λ^(n+1),的
a[n+1]/λ^(n+1)=a[n]/λ^n+1+[(2-λ)/λ]×2^n
构造新数列{b[n]},令b[n]=a[n]/λ^n,b[1]=2/λ
即 b[n+1]=b[n]+1+[(2-λ)/λ]×(2/λ)^n
b[n]=b[n-1]+1+[(2-λ)/λ]×(2/λ)^(n-1）
∶
∶
b[3]=b[2]+1+[(2-λ)/λ]×(2/λ)^2
b[2]=b[1]+1+[(2-λ)/λ]×2/λ
各式相加,得
b[n]=b[1]+n-1+[(2-λ)/λ]×∑[1≤k≤n](2/λ)^(k-1）
b[n]=2/λ+n-1+(2/λ)^n-2/λ
即 b[n]=n-1+(2/λ)^n
从而所求的通项公式:a[n]=[n-1+(2/λ)^n]×λ^n=(n-1)λ^n+2^n
s[n]=2+(λˆ2+2ˆ2）+(2λˆ3+2ˆ3)+(3λˆ4+2ˆ4)+…+(n-1)λ^n+2^n
=(λˆ2+2λˆ3+3λˆ4+…+(n-1)λ^n)+(2+2ˆ2+2ˆ3+2ˆ4+…+2ˆn)
令c[n]=λˆ2+2λˆ3+3λˆ4+…+(n-1)λ^n,d[n]=2+2ˆ2+2ˆ3+2ˆ4+…+2ˆn
则 s[n]=c[n]+d[n]
又 c[n]=λˆ2+2λˆ3+3λˆ4+…+(n-1)λ^n
λc[n]=λˆ3+2λˆ4+3λˆ5+…+(n-1)λ^(n+1)
错位相减,得
(1-λ)c[n]=λˆ2+λˆ3+λˆ4+…+λ^n-(n-1)λ^(n+1)
(1-λ)c[n]=λˆ2[1-λˆ(n-1)]/(1-λ)-(n-1)λ^(n+1)
所以 c[n]=λˆ2[1-λˆ(n-1)]/(1-λ)ˆ2-(n-1)λ^(n+1)/(1-λ)
d[n]=2+2ˆ2+2ˆ3+…+2ˆn=2ˆ(n+1)-2
所以前n项和s[n]=λˆ2[1-λˆ(n-1)]/(1-λ)ˆ2-(n-1)λ^(n+1)/(1-λ)+2ˆ(n+1)-2