Question:

Expand using Binomial Theorem (1+X22X)4,X0( 1+ \frac{X}{2} - \frac{2}{X})^4, X≠0

Updated On: Oct 26, 2023
Hide Solution
collegedunia
Verified By Collegedunia

Solution and Explanation

Using Binomial Theorem, the given expression (1+x22x)4( 1 +\frac{ x}{2} - \frac{2}{x})^4 can be expanded as
[(1+x2)2x]4[ ( 1 +\frac{ x}{2} )- \frac{2}{x} ]^4

= 4C0(1+x2)4 4C1(1+x2)3(2x)+ 4C2(1+x2)2(2x)2 4C3(1+x2)(2x)3+ 4C4(2x)4=\space^ 4C_0(1 + \frac{x}{2})^4 - \space^4C_1(1 +\frac{ x}{2})^3(\frac{2}{x}) + \space^4C_2(1 +\frac{ x}{2})^2(\frac{2}{x})^2 - \space^4C_3(1 +\frac{ x}{2}) (\frac{2}{x})^3 + \space^4C_4(\frac{2}{x})^4

=(1+x2)44(1+x2)3(2x)+6(1+x+x24)(4x2)4(1+x2)(8x3)+16x4= (1 +\frac{ x}{2})^4 -4(1+\frac{x}{2})^3(\frac{2}{x}) + 6(1+x+ \frac{x^2}{4}) (\frac{4}{x^2})- 4(1 + \frac{x}{2})(\frac{8}{x^3})+\frac{16}{x^4}

= (1+x2)4 8x((1+x2)3+24x2+24x+632x316x2+16x4= (1 + \frac{x}{2})^4 - \frac{8}{x}((1 +\frac{ x}{2})^3 + \frac{24}{x^2 }+\frac{ 24}{x }+6 - \frac{32}{x^3} -\frac{16}{x^2} +\frac{ 16}{x^4}

= (1+x2)4 8x((1+x2)3+8x2+24x+6+32x3+16x4           ......(1)= (1 + \frac{x}{2})^4 - \frac{8}{x}((1 +\frac{ x}{2})^3 + \frac{8}{x^2} +\frac{24}{x }+6 + \frac{32}{x^3} +\frac{16}{x^4}                      ......(1)

Again by using Binomial Theorem, we obtain

(1+x2)4= 3C0(1)4+ 4C1(1)3(x2)+ 4C2(1)2(x2)2 + 4C3(1)(x2)3 + 4C4(x2)4(1 +\frac{ x}{2})^4 = \space^3C_0 (1)^4+\space^ 4C_1(1)^3(\frac{x}{2})+ \space ^4C_2(1)^2(\frac{x}{2})^2 + \space^4C_3(1)(\frac{x}{2})^3 + \space^4C_4(\frac{x}{2})^4 =1+4×x2+6×x24+4×x38+x416   ...(2)=1+4 \times\frac{x}{2}+6\times \frac{x^2}{4} + 4 \times \frac{x^3}{8} +\frac{x^4}{16}     ...(2)

(1+x2)3= 3C0(1+x2)3 3C1(1)2(x2)+ 3C2(1)(x2)2 3C3(x2)3(1 + \frac{x}{2})^3= \space^3C_0(1 +\frac{ x}{2})^3 - \space^3C_1(1 )^2 (\frac{x}{2})+\space^ 3C_2(1) (\frac{x}{2})^2 - \space^3C_3( \frac{x}{2})^3 =1+3x2+3x24+x38   ....(3)=1 + \frac{3x}{2} + \frac{3x^2}{4} + \frac{x^3}{8}     ....(3)

from (1), (2) and (3), we obtain
[(1+x2)2x]4[(1 + \frac{x}{2}) - \frac{2}{x}]^4

=1+2x+3x22+x32+x4168x(1+3x2+3x24+x38)+8x2+24x+632x3+16x4= 1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} +\frac{ x^4}{16} -\frac{ 8}{x }(1 + \frac{3x}{2} +\frac{ 3x^2}{4} +\frac{ x^3}{8}) + \frac{8}{x^2} + \frac{24}{x} + 6 - \frac{32}{x^3} + \frac{16}{x^4}

= 1+2x+32x2+x32+x4168x126xx2+8x2+24x+632x3+16x4= 1 + 2x + \frac{3}{2}x^2 + \frac{x^3}{2} + \frac{x^4}{16} - \frac{8}{x} -12 - 6x - x^2 + \frac{8}{x^2} + \frac{24}{x} + 6 - \frac{32}{x^3} + \frac{16}{x^4}

=16x+8x232x3+16x44x=x22+x32+x4165= \frac{16}{x} + \frac{8}{x^2} - \frac{32}{x^3} + \frac{16}{x^4} - 4x = \frac{x^2}{2} +\frac{ x^3}{2} +\frac{ x^4}{16} -5

Was this answer helpful?
0
0

Questions Asked in CBSE Class XI exam

View More Questions

Concepts Used:

Binomial Theorem

The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is 

Properties of Binomial Theorem

  • The number of coefficients in the binomial expansion of (x + y)n is equal to (n + 1).
  • There are (n+1) terms in the expansion of (x+y)n.
  • The first and the last terms are xn and yn respectively.
  • From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
  • The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.