问题
解答题
已知数列{an}中，a1=1，an＜an+1，设bn=
（Ⅰ）bn＜2(
（Ⅱ）若数列{an}是公比为q且q≥3的等比数列，则Sn＜1． |
答案
证明：（Ⅰ）由题意可知an＞0
∵bn-2(
-1 an
)1 an+1
=
-2(an+1-an an+1• an+1
-1 an
)1 an+1
=
-2an+1-an an+1 an+1
-an+1 an
•an+1 an
=
．(
-an+1
)(an
•an+1
+an-2an+1)an an+1•
•an+1 an
又an＜an+1，∴
-an+1
＞0，an
•an+1
＜an+1，an
•an+1
+an-2an+1＜0，an
则
＜0．(
-an+1
)(an
•an+1
+an-2an+1)an an+1•
•an+1 an
∴bn＜2(
-1 an
)；1 an+1
（Ⅱ）数列{an}是首项a1=1，公比为q且q≥3的等比数列，
∴an=a1qn-1=qn-1．
∴bn=
=an+1-an an+1• an+1
=q-qn-qn-1 q 3n 2
(1-q-1)．n 2
Sn=b1+b2+…+bn
=(1-q-1)(q-
+q-1 2
+q-2 2
+…+q-3 2
)n 2
=(1-q-1)•q-
(1-q-1 2
)n 2 1-q- 1 2
=(
+1 q
)(1-1 q
)．1 qn
∵q≥3，∴0＜
+1 q
≤1 q
+1 3
=1 3
＜1．
+13 3
0＜1-
＜1．1 qn
∴Sn=(
+1 q
)(1-1 q
)＜1．1 qn