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MAA AMC
AMC 12/AHSME
2021 AMC 12/AHSME Fall
18
18
Part of
2021 AMC 12/AHSME Fall
Problems
(1)
It's not like I'm approaching you or anything, baka!
Source: 12B #18
11/17/2021
Set
u
0
=
1
4
,
u_0 = \frac{1}{4},
u
0
=
4
1
,
and for
k
≥
0
k \geq 0
k
≥
0
let
u
k
+
1
u_{k+1}
u
k
+
1
be determined by the recurrence
u
k
+
1
=
2
u
k
−
2
u
k
2
.
u_{k+1} = 2u_k - 2u_k^2.
u
k
+
1
=
2
u
k
−
2
u
k
2
.
This sequence tends to a limit, call it
L
.
L.
L
.
What is the least value of
k
k
k
such that
∣
u
k
−
L
∣
≤
1
2
1000
?
|u_k - L| \leq \frac{1}{2^{1000}}?
∣
u
k
−
L
∣
≤
2
1000
1
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
10
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
97
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
253
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
329
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
401
<span class='latex-bold'>(A)</span>\ 10 \qquad<span class='latex-bold'>(B)</span>\ 97 \qquad<span class='latex-bold'>(C)</span>\ 253 \qquad<span class='latex-bold'>(D)</span>\ 329 \qquad<span class='latex-bold'>(E)</span>\ 401
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
10
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
97
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
253
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
329
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
401
AMC
AMC 12
AMC 12 B