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Problem
Problem
If
∑
k
=
1
n
f
(
a
+
k
)
=
16
(
2
n
−
1
)
\sum_{k=1}^n f(a+k) = 16(2^n - 1)
∑
k
=
1
n
f
(
a
+
k
)
=
16
(
2
n
−
1
)
, where
f
(
x
+
y
)
=
f
(
x
)
f
(
y
)
f(x+y) = f(x)f(y)
f
(
x
+
y
)
=
f
(
x
)
f
(
y
)
for all
x
,
y
∈
N
,
f
(
1
)
=
2
x, y \in N, f(1) = 2
x
,
y
∈
N
,
f
(
1
)
=
2
, then
f
(
a
)
=
f(a) =
f
(
a
)
=
(a) 8 (b) 6 (c) 4 (d) 2
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