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Mathematics
Mathematics
Problem
The Maclaurin series of
p
(
x
)
=
ln
(
1
+
k
x
)
p(x)=\ln(1+kx)
p
(
x
)
=
ln
(
1
+
k
x
)
is
ln
(
1
+
k
x
)
=
k
x
−
k
2
x
2
2
+
k
3
x
3
3
−
⋯
.
\ln(1+kx)=kx-\frac{k^{2}x^{2}}{2}+\frac{k^{3}x^{3}}{3}-\cdots.
ln
(
1
+
k
x
)
=
k
x
−
2
k
2
x
2
+
3
k
3
x
3
−
⋯
.
If the coefficient of
x
2
x^{2}
x
2
in this series is
−
9
2
-\dfrac{9}{2}
−
2
9
and
k
>
0
k>0
k
>
0
, what is the value of
k
k
k
? a)
9
9
9
b)
3
3
3
c)
3
2
\dfrac{3}{2}
2
3
d)
9
2
\sqrt{\dfrac{9}{2}}
2
9
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