问题
选择题
在等差数列{an}中,a2=4,a6=12,,那么数列{
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答案
∵等差数列{an}中,a2=4,a6=12;
∴公差d=
=a6- a2 6-2
=2;12-4 6-2
∴an=a2+(n-2)×2=2n;
∴
=an 2n+1
;n 2n
∴
的前n项和,an 2n+1
Sn=1×
+2×(1 2
)2+3×(1 2
)3+…+(n-1)×(1 2
)n-1+n×(1 2
)n1 2
Sn=1×(1 2
)2+2×(1 2
)3+3×(1 2
)4…+(n-1)×(1 2
)n+n×(1 2
)n+11 2
两式相减得
Sn=1 2
+(1 2
)2+(1 2
)3+…+(1 2
)n-n(1 2
)n+11 2
=
- n(
-(1 2
)n+11 2 1- 1 2
)n+11 2
∴Sn=1+n+1 2n
故选B