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由a1,a4,a13成等比数列,得a42=a1a13,
即(a1+3d)2=a1(a1+12d),所以a12+6a1d+9d2=a12+12a1d.
9d2=6a1d,a1=
d.则d=
a1=
×3=2.
(Ⅰ)an=a1+(n-1)d=3+2(n-1)=2n+1;
(Ⅱ)Sn=na1+
=3n+n2?n=n(n+2).
则
=
=
(
?
),
所以
+
+…
=
(1?
+
?
+
?
+…+
?
+
?
)
=
(1+
?
?
)=
?<
即(a1+3d)2=a1(a1+12d),所以a12+6a1d+9d2=a12+12a1d.
9d2=6a1d,a1=
3 |
2 |
2 |
3 |
2 |
3 |
(Ⅰ)an=a1+(n-1)d=3+2(n-1)=2n+1;
(Ⅱ)Sn=na1+
n(n?1)d |
2 |
则
1 |
Sn |
1 |
n(n+2) |
1 |
2 |
1 |
n |
1 |
n+2 |
所以
1 |
S1 |
1 |
S2 |
1 |
Sn |
1 |
2 |
1 |
3 |
1 |
2 |
1 |
4 |
1 |
3 |
1 |
5 |
1 |
n?1 |
1 |
n+1 |
1 |
n |
1 |
n+2 |
=
1 |
2 |
1 |
2 |
1 |
n+1 |
1 |
n+2 |
3 |
4 |
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