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MAA AMC
AMC 12/AHSME
1971 AMC 12/AHSME
7
Simplify
Simplify
Source: 1971 AHSME Problem 7
April 18, 2014
AMC
Problem Statement
2
−
(
2
k
+
1
)
−
2
−
(
2
k
−
1
)
+
2
−
2
k
2^{-(2k+1)}-2^{-(2k-1)}+2^{-2k}
2
−
(
2
k
+
1
)
−
2
−
(
2
k
−
1
)
+
2
−
2
k
is equal to
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
2
−
2
k
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2
−
(
2
k
−
1
)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
−
2
−
(
2
k
+
1
)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
2
<span class='latex-bold'>(A) </span>2^{-2k}\qquad<span class='latex-bold'>(B) </span>2^{-(2k-1)}\qquad<span class='latex-bold'>(C) </span>-2^{-(2k+1)}\qquad<span class='latex-bold'>(D) </span>0\qquad <span class='latex-bold'>(E) </span>2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
2
−
2
k
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2
−
(
2
k
−
1
)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
−
2
−
(
2
k
+
1
)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
2
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