NCERT Solutions
Class 11 Maths
Binomial Theorem

Ex.Misc.Q.9
Expand using Binomial Theorem (1 + –
)4, x ≠ 0.
Using Binomial Theorem, the given expression (1 + –
)4 can be expanded as
(1 + –
)4 = {(1 +
) –
}4
= 4C0 (1 + )4 - 4C1 (1 +
)3 (
) + 4C2 (1 +
)2 (
)2 - 4C3 (1 +
)(
)3 + 4C4 (
)4
= (1 + )4 - 4(1 +
)3 (
) + 6(1 +
)2 (4 ÷ x2) - 4(1 +
)(8 ÷ x3) + (16 ÷ x4)
= (1 + )4 - (1 + x2
)3 (8x
) + (1 + x + (x2 ÷ 4)) (24 ÷ x2) - (1 +
)(32 ÷ x3) + (16 ÷ x4)
= (1 + )4 - (1 + x2 )3 (
) + (24 ÷ x2) +
+ 6 – (32 ÷ x3) – (16 ÷ x2) + (16 ÷ x4)
= (1 + )4 - (1 +
)3 (
) + (8 ÷ x2) + (24 ÷ x) + 6 – (32 ÷ x3) + (16 ÷ x4)
=> (1 + –
)4 = (1 +
)4 - (1 +
)3 (
) + (8 ÷ x2) +
+ 6 – (32 ÷ x3) + (16 ÷ x4) ………. (1)
Again, by using Binomial Theorem, we obtain
(1 + )4 = 4C0 (1)4 + 4C1 (1)3 (
) + 4C2 (1)2 (
)2 + 4C3 (1) (
)3 + 4C4 (
)4
= 1 + 4 × (x2 ) + 6 × (x2 ÷ 4) + 4 × (x3 ÷ 8) + (x4 ÷ 16)
=> (1 + )4 = 1 + 2x + (3x2 ÷ 2) + (x3 ÷ 2) + (x4 ÷ 16) …………... (2)
(1 + )3 = 3C0 (1)3 + 3C1 (1)2 (
) + 3C2 (1) (
)2 + 3C3 (
)3
=> (1 + )3 = 1 +
+ 3(x2 ÷ 4) + (x3 ÷ 8) …………... (3)
From equation (1), (2), and (3), we get
(1 + –
)4 = 1 + 4 × (
) + 6 × (x2 ÷ 4) + 4 × (x3 ÷ 8) + (x4 ÷ 16) –
[1 + + (3x2 ÷ 4) + (x3 ÷ 8)] (
) + (8 ÷ x2) + (24 ÷ x) + 6 – (32 ÷ x3) + (16 ÷ x4)
= 1 + 2x + x2 + (x3 ÷ 2) + (x4 ÷ 16) –
- 12 – 6x – x2 + (8 ÷ x2) + (24 ÷ x) + 6 – (32 ÷ x3) + (16 ÷ x4)
= + (8 ÷ x2) – (32 ÷ x3) +(16 ÷ x4) + (x2 ÷ 2) + (x3 ÷ 2) + (x4 ÷ 16) - 5