题目内容
设数列{an}满足a1=6,a2=4,a3=3,且数列{an+1-an}(n∈N*)是等差数列,求数列{an}的通项公式.
∵a1=6,a2=4,a3=3,
∴a2-a1=-2,a3-a2=-1,且-1-(-2)=1,
数列{an+1-an}是-2为首项,1为公差的等差数列,
∴an+1-an=-2+(n-1)×1=n-3,
∴an=(an-an-1)+(an-1-an-2)+(an-2-an-3)+…+(a2-a1)+a1
=(n-4)+(n-5)+(n-6)+…+(-2)+6
=
+6=
n2-
n+9
∴a2-a1=-2,a3-a2=-1,且-1-(-2)=1,
数列{an+1-an}是-2为首项,1为公差的等差数列,
∴an+1-an=-2+(n-1)×1=n-3,
∴an=(an-an-1)+(an-1-an-2)+(an-2-an-3)+…+(a2-a1)+a1
=(n-4)+(n-5)+(n-6)+…+(-2)+6
=
(n-1)(n-4-2) |
2 |
1 |
2 |
7 |
2 |
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