(2x-1/x)^6 expand using binomial theorem
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Answered by
0
Answer:
Use the binomial expansion theorem to find each term. The binomial theorem states
(
a
+
b
)
n
=
n
∑
k
=
0
n
C
k
⋅
(
a
n
−
k
b
k
)
(a+b)n=∑k=0nnCk⋅(an-kbk).
6
∑
k
=
0
6
!
(
6
−
k
)
!
k
!
⋅
(
2
x
)
6
−
k
⋅
(
−
1
x
)
k
∑k=066!(6-k)!k!⋅(2x)6-k⋅(-1x)k
Expand the summation.
6
!
(
6
−
0
)
!
0
!
⋅
(
2
x
)
6
−
0
⋅
(
−
1
x
)
0
+
6
!
(
6
−
1
)
!
1
!
⋅
(
2
x
)
6
−
1
⋅
(
−
1
x
)
+
…
+
6
!
(
6
−
6
)
!
6
!
⋅
(
2
x
)
6
−
6
⋅
(
−
1
x
)
6
6!(6-0)!0!⋅(2x)6-0⋅(-1x)0+6!(6-1)!1!⋅(2x)6-1⋅(-1x)+…+6!(6-6)!6!⋅(2x)6-6⋅(-1x)6
Simplify the exponents for each term of the expansion.
1
⋅
(
2
x
)
6
⋅
(
−
1
x
)
0
+
6
⋅
(
2
x
)
5
⋅
(
−
1
x
)
+
…
+
1
⋅
(
2
x
)
0
⋅
(
−
1
x
)
6
1⋅(2x)6⋅(-1x)0+6⋅(2x)5⋅(-1x)+…+1⋅(2x)0⋅(-1x)6
Simplify the polynomial result.
64x6−192x4+240x2−160+60x2−12x4+1x6
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Answer:
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