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Mathematics
Problem
The inverse-tangent function satisfies
arctan
x
=
∑
n
=
0
∞
(
−
1
)
n
x
2
n
+
1
2
n
+
1
\arctan x =\sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{2n+1}
arctan
x
=
∑
n
=
0
∞
(
−
1
)
n
2
n
+
1
x
2
n
+
1
for
∣
x
∣
≤
1
|x|\le 1
∣
x
∣
≤
1
. Which of the following gives a power-series representation for
1
1
+
x
2
\dfrac{1}{1+x^{2}}
1
+
x
2
1
valid for
∣
x
∣
<
1
|x|<1
∣
x
∣
<
1
? a)
∑
n
=
0
∞
(
−
1
)
n
x
n
\displaystyle \sum_{n=0}^{\infty} (-1)^n x^{n}
n
=
0
∑
∞
(
−
1
)
n
x
n
b)
∑
n
=
0
∞
(
−
1
)
n
(
2
n
+
1
)
x
2
n
\displaystyle \sum_{n=0}^{\infty} (-1)^n (2n+1) x^{2n}
n
=
0
∑
∞
(
−
1
)
n
(
2
n
+
1
)
x
2
n
c)
∑
n
=
0
∞
(
−
1
)
n
x
2
n
2
n
+
1
\displaystyle \sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n}}{2n+1}
n
=
0
∑
∞
(
−
1
)
n
2
n
+
1
x
2
n
d)
∑
n
=
0
∞
(
−
1
)
n
x
2
n
\displaystyle \sum_{n=0}^{\infty} (-1)^n x^{2n}
n
=
0
∑
∞
(
−
1
)
n
x
2
n
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