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设数列{an}满足a1+2a2=3,且对任意的n∈N*,点Pn(n,an)都有PnPn+1=(1,2),则{an}的前n项和Sn为()A.n(n−43)B.n(n−34)C.n(n−23)D.n(n−12)

题目详情
设数列{an}满足a1+2a2=3,且对任意的n∈N*,点Pn(n,an)都有
PnPn+1
=(1,2),则{an}的前n项和Sn为(  )
A. n(n−
4
3
)
B. n(n−
3
4
)
C. n(n−
2
3
)
D. n(n−
1
2
)
▼优质解答
答案和解析
∵Pn(n,an),∴Pn+1(n+1,an+1),故
PnPn+1
=(1,an+1−an)  =(1,2)
an+1-an=2,∴an是等差数列,公差d=2,将a2=a1+2,代入a1+2a2=3中,
解得a1=−
1
3
,∴an=−
1
3
+2(n−1)=2n−
7
3

Sn=
a1+an
2
n=
1
3
+2n−
7
3
2
n=(n−
4
3
)n,
故选A.