已知等差数列{a n }的首项a 1 =1,公差d=1,前n项和为S n , b n = 1 S n ,(1)求

2024-12-19 14:48:54
推荐回答(1个)
回答1:

(1)∵等差数列{a n }中a 1 =1,公差d=1
S n =n a 1 +
n(n-1)
2
d=
n 2 +n
2

b n =
2
n 2 +n
…(4分)
(2)∵ b n =
2
n 2 +n
=
2
n(n+1)
…(6分)
b 1 + b 2 + b 3 +…+ b n =2(
1
1×2
+
1
2×3
+
1
3×4
+…+
1
n(n+1)
)

= 2(1-
1
2
+
1
2
-
1
3
+
1
3
-
1
4
+…+
1
n
-
1
n+1
)
…(8分)
= 2(1-
1
n+1
)
…(11分)
∵n>0,
0<
1
n+1
<1

0<2(1-
1
n+1
)<2

∴b 1 +b 2 +…+b n <2.            …(14分)