已知数列{an}中、a1=1, a2=4 (an+2)=(2an+1) - (an) +2,求通项公式an
an+2-an+1=an+1-an+2
令bn+2=an+2-an+1
则b2=a2-a1=3
bn=bn-1+2
bn-1=bn-2+2
.
b3=b2+2
b2=3
叠加上式得:
bn=3+2(n-2)=an-an-1
an= an-1+3+2(n-2)
an-1=an-2+3+2(n-3)
a2=a1+3+2(1-1)
叠加得:
an=a1+3(n-1)+2(0+1+2+…+n-3+n-2)
=1+3(n-1)+(n-1)(n-2)
=1+(n-1)(n+1)
=n2