已知等差数列{a n }的首项a 1 =3,公差d≠0,其前n项和为S n ,且a 1 ,a 4 ,a 13 成等比数列.(Ⅰ)求

2025-01-18 16:09:31
推荐回答(1个)
回答1:

由a 1 ,a 4 ,a 13 成等比数列,得 a 4 2 = a 1 a 13
( a 1 +3d ) 2 = a 1 ( a 1 +12d) ,所以 a 1 2 +6 a 1 d+9 d 2 = a 1 2 +12 a 1 d
9d 2 =6a 1 d, a 1 =
3
2
d
.则 d=
2
3
a 1 =
2
3
×3=2

(Ⅰ)a n =a 1 +(n-1)d=3+2(n-1)=2n+1;
(Ⅱ) S n =n a 1 +
n(n-1)d
2
=3n+ n 2 -n=n(n+2)

1
S n
=
1
n(n+2)
=
1
2
(
1
n
-
1
n+2
)

所以
1
S 1
+
1
   S 2
+…
1
S n
=
1
2
(1-
1
3
+
1
2
-
1
4
+
1
3
-
1
5
+…+
1
n-1
-
1
n+1
+
1
n
-
1
n+2
)

=
1
2
(1+
1
2
-
1
n+1
-
1
n+2
)=
3
4
-
1
2(n+1)
-
1
2(n+2)
3
4