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NCERT Class XI Mathematics - Binomial Theorem - Solutions

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Question : 35
Total: 36
Expand using Binomial Theorem (1+
x
2
−
2
x
)4 , x ≠ 0
Solution:  
Let
x
2
−
2
x
 = y
∴ (1+
x
2
−
2
x
)4 = (1+y)4
=
4
‌
C0+
4
‌
C1y+
4
‌
C3y3+
4
‌
C4y4
=
4
‌
C0+
4
‌
C1[
x
2
−
2
x
]+
4
‌
C2[
x
2
−
2
x
]2 +
4
‌
C3[
x
2
−
2
x
]3+
4
‌
C4[
x
2
−
2
x
]4
= 1 + 4 [
x
2
−
2
x
] +
6[
2
‌
C0(
x
2
)2+
2
‌
C1(
x
2
)(−
2
x
)+
2
‌
C2(−
2
x
)2]
+
4[
3
‌
C0(
x
2
)3+
3
‌
C1(
x
2
)2(−
2
x
)+
3
‌
C2(
x
2
)(−
2
x
)2+
3
‌
C3(−
2
x
)3]
+
[
4
‌
C0(
x
2
)4+
4
‌
C1(
x
2
)3(−
2
x
)+
4
‌
C2(
x
2
)2(−
2
x
)2+
4
‌
C3(
x
2
)(−
2
x
)3+
4
‌
C4(−
2
x
)4]

= 1 + 2x -
8
x
+6[
x2
4
−2+
4
x2
] + 4[
x3
8
−
3x
2
+
6
x
−
8
x3
] + [
x4
16
−x2+6−
16
x2
+
16
x4
]
= 1 + 2x -
8
x
+
3
2
x2 - 12 +
24
x2
+
x3
2
 - 6x +
34
x
−
32
x3
+
x4
16
−x2 + 6 -
16
x2
+
16
x4
=
16
x4
−
32
x3
+
8
x2
+
16
x
 - 5 - 4x +
x2
2
+
x3
2
+
x4
16
=
16
x
+
8
x2
−
32
x3
+
16
x4
 - 4x +
x2
2
+
x3
2
+
x4
16
−5
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