数列{an}满足an+2-2an+1+an=0,a3=11,前9项和为153 求an
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数列{an}满足an+2-2an+1+an=0,a3=11,前9项和为153 求an
数列{an}满足an+2-2an+1+an=0,a3=11,前9项和为153 求an
数列{an}满足an+2-2an+1+an=0,a3=11,前9项和为153 求an
an+2-2an+1+an=0
所以:an+2+an=2an+1
所以:{an}是一个等差数列.
s9==9a5==153
a5==17==a3+2d
a3=11
d==3
a1=5
an=5+(n-1)d=5+(n-1)*3=3n+2
a(n+2)-2a(n+1)+an=0
故a(n+2)-a(n+1)=a(n+1)-an
即an是等差数列
设公差为3
故a3=a1+2d=11
S9=9a1+36d=153
解之a1=5,d=3
故an=5+(n-1)3=3n+2
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