已知數(shù)列{an}滿足a1=1,a2=-13,an+2-2an+1+an=2n-6
b1=a2-a1
=-13-1
=-14
bn=a(n+1)-an
a(n+2)-2a(n+1)+an=2n-6
a(n+2)-a(n+1)-a(n+1)+an=2n-6
a(n+2)-a(n+1)-[a(n+1)+an]=2n-6
b(n+1)-bn=2n-6
所以bn-b(n-1)=2(n-1)-6
bn-b(n-1)=2(n-1)-6
.
b3-b2=2*2-6
b2-b1=2*1-6
以上等式相加得
bn-b1=2*1-6+2*2-6+.+2(n-1)-6
bn-b1=2*(1+2+3+.+n-1)-6(n-1)
bn-b1=n(n-1)-6n+6
bn-(-14)=n^2-n-6n+6
bn+14=n^2-7n+6
bn=n^2-7n-8
bn-b1=2*1-6+2*2-6+.+2(n-1)-6
bn-b1=2*(1+2+3+.+n-1)-6(n-1)
bn-b1=n(n-1)-6n+6
2為什么約掉了.
已知數(shù)列{an}滿足a1=1,a2=-13,a(n+2)-2(an+1)+an=2n-6 1)設(shè)bn=A(n+1)-An,求數(shù)列{bn}的通項公式