问题
填空题
已知数列{an}的通项公式是an=1-
|
答案
∵an=1-
,1 2n
∴Sn=(1-
)+(1-1 2
)+…+(1-1 22
)1 2n
=(1+1+…+1)-(
-1 2
-…-1 22
)1 2n
=n-
=n-1+
(1-1 2
)1 2n 1- 1 2 1 2n
令n-1+
=1 2n
,解之可得n=6321 64
故答案为:6
已知数列{an}的通项公式是an=1-
|
∵an=1-
,1 2n
∴Sn=(1-
)+(1-1 2
)+…+(1-1 22
)1 2n
=(1+1+…+1)-(
-1 2
-…-1 22
)1 2n
=n-
=n-1+
(1-1 2
)1 2n 1- 1 2 1 2n
令n-1+
=1 2n
,解之可得n=6321 64
故答案为:6