Question

已知数列{an}满足a1=1,a2=-13,an+2-2an+1=2n-6设bn=an+1 -an求

Answer 1

1.
bn=a(n+1)-an,b(n+1)=a(n+2)-a(n+1)
b1=a2-a1=-14
a(n+2)-2a(n+1)+an=2n-6
即[a(n+2)-a(n+1)]-[a(n+1)-an]=2n-6
即b(n+1)-bn=2n-6
bn-b(n-1)=2(n-1)-6
b(n-1)-b(n-2)=2(n-2)-6
b(n-2)-b(n-3)=2(n-3)-6
.
.
.
b3-b2=2*2-6
b2-b1=2*1-6
全加得
bn-b1=2[1+2+...+ (n-2)+(n-1)]-6(n-1)
=2*n(n-1)/2-6(n-1)
=n��-7n+6
bn=n��-7n+6-14=n��-7n-8