Q.22. Expand (2x - 1/x)^6
Answers
And: (
2
x
−
1
x
)
6
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
)
.
6
∑
k
=
0
6
!
(
6
−
k
)
!
k
!
⋅
(
2
x
)
6
−
k
⋅
(
−
1
x
)
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
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
Simplify the polynomial result.
64
x
6
−
192
x
4
+
240
x
2
−
160
+
60
x
2
−
12
x
4
+
1
x
6
Answer:
Step-by-step explanation:
(a+b)^6 = 6C0 (a^6).(b^0) + 6C1 (a^5).(b^1) + 6C2 (a^4).(b^2) + 6C3 (a^3).(b^3) + 6C4 (a^2).(b^4) + 6C5 (a^1).(b^5) + 6C6 (a^0).(b^6)
(2x - 1/x)^6 = [(2x)^6] + [6 . (2x)^5 . (1/x)] + [15 . (2x)^4 . (1/x)^2] + [20 . (2x)^3 . (1/x)^3] + [15 . (2x)^2 . (1/x)^4] + [6 . (2x) . (1/x)^5] + [(1/x)^6]
Hope this helps you...