Answer :
Step-by-step explanation:
Explanation:
The idea is to relate this expression to the known power series expansion
1
1
−
x
=
∞
∑
n
=
0
x
n
Temporarily disregard the
x
2
and consider
f
(
x
)
=
x
2
1
(
1
−
2
x
)
2
.
Take the integral of
1
(
1
−
2
x
)
2
:
∫
d
x
(
1
−
2
x
)
2
Quick substitution:
u
=
1
−
2
x
d
u
=
−
2
d
x
,
−
1
2
d
u
=
d
x
−
1
2
∫
u
−
2
d
u
=
1
2
u
=
1
2
(
1
−
2
x
)
Thus, knowing that differentiating this integrated expression returns the original
1
(
1
−
2
x
)
2
,
we can say
f
(
x
)
=
x
2
d
d
x
1
2
(
1
−
2
x
)
Let's find the power series representation for the differentiated expression:
f
(
x
)
=
x
2
d
d
x
1
2
⋅
1
1
−
2
x
We can easily relate
1
1
−
2
x
to
1
1
−
x
=
∞
∑
n
=
0
x
n
:
1
1
−
2
x
=
∞
∑
n
=
0
(
2
x
)
n
=
∞
∑
n
=
0
2
n
x
n
So,
f
(
x
)
=
x
2
d
d
x
1
2
∞
∑
n
=
0
2
n
x
n
We can absorb the
1
2
in:
f
(
x
)
=
x
2
d
d
x
∞
∑
n
=
0
2
n
−
1
x
n
Differentiate the summation with respect to
x
, recalling that differentiating the summation causes the index to shift up by
1
:
f
(
x
)
=
x
2
∞
∑
n
=
0
d
d
x
2
n
−
1
x
n
f
(
x
)
=
x
2
∞
∑
n
=
1
2
n
−
1
n
x
n
−
1
Multiply in the
x
2
:
f
(
x
)
=
∞
∑
n
=
1
2
n
−
1
n
x
n
−
1
+
2
f
(
x
)
=
∞
∑
n
=
1
2
n
−
1
n
x
n
+
1
