(1)
a2=2*(-3)+2^2+3=1
a3=2*1+2^3+3=13
(2)
an+3=2(an-1+3)+2^n
兩邊同時(shí)除以2^n
(an+3)/2^n=(an-1+3)/2^(n-1)+1
即bn=bn-1+1
b1=(a1+3)/2=0
所以bn=n-1,bn是等差數(shù)列
(3)
bn=(an+3)/2^n
n-1=(an+3)/2^n
an=(n-1)2^n-3=n·2^n-2^n-3
Sn=∑n·2^n-∑2^n-n·3
∑2^n=2+2²+2³+...+2^n=2^(n+1)-2
∑n·2^n=1×2+2×4+3×8+...+(n-1)·2^(n-1)+n·2^n -------①
2∑n·2^n=1×4+2×8+3×16+...+(n-1)·2^n+n·2^(n+1)-------②
兩式相減得
∑n·2^n=n·2^(n+1)-(2^n+2^(n-1)+...+8+4+2)=n·2^(n+1)-∑2^n
∴Sn=∑n·2^n-∑2^n-n·3
=n·2^(n+1)-2∑2^n-3n
=n·2^(n+1)-2*[2^(n+1)-2]-3n
=(n-2)2^(n+1)-3n+4
所以Sn=(n-2)2^(n+1)-3n+4