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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
=(1,2),则{an}的前n项和Sn为( )
A. n(n−
)
B. n(n−
)
C. n(n−
)
D. n(n−
)
PnPn+1 |
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
=(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=−
,∴an=−
+2(n−1)=2n−
∴Sn=
n=
n=(n−
)n,
故选A.
PnPn+1 |
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 |
1 |
3 |
7 |
3 |
∴Sn=
a1+an |
2 |
−
| ||||
2 |
4 |
3 |
故选A.
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