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  • Expand using Binomial Theorem ( 1 + x / 2 - 2 / x) ^4, x not equal to 0

Q9.    Expand using Binomial Theorem \left(1 + \frac{x}{2} - \frac{2}{x} \right )^4, \ x\neq 0

Answers (1)

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Given the expression,

\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4, \ x\neq 0

Binomial expansion of this expression is 

\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4\\=\left ( \left (1 + \frac{x}{2} \right ) -\frac{2}{x}\right )^4=^4C_0\left ( 1+\frac{x}{2} \right )^4-^4C_1\left ( 1+\frac{x}{2} \right )^3\left ( \frac{2}{x} \right )+^4C_2\left ( 1+\frac{x}{2} \right )^2\left ( \frac{2}{x} \right )^2-^4C_3\left ( 1+\frac{x}{2} \right )\left ( \frac{2}{x} \right )^3+^4C_4\left ( \frac{2}{x} \right )^4

\Rightarrow \left ( 1+\frac{x}{2} \right )^4-\frac{8}{x}\left ( 1+\frac{x}{2} \right )^3+\frac{24}{x^2}+\frac{24}{x}+6-\frac{32}{x^3}+\frac{16}{x^4}..........(1)

Now Applying Binomial Theorem again,

\left ( 1+\frac{x}{2} \right )^4=^4C_0(1)^4+^4C_1(1)^3\left ( \frac{x}{2} \right )+^4C_2(1)^2\left ( \frac{x}{2} \right )^2+^4C_3(1)\left ( \frac{x}{2} \right )^3+^4C_4\left ( \frac{x}{2} \right )^4

= 1+ 4\left ( \frac{x}{2} \right )+6\left ( \frac{x^2}{4} \right )+4\left ( \frac{x^3}{8} \right )+\frac{x^4}{16}

=1+2x+\frac{3x^2}{2}+\frac{x^3}{3}+\frac{x^4}{16}..............(2)

And 

\left ( 1+\frac{x}{2} \right )^3=^3C_0(1)^3+^3C_1(1)^2\left ( \frac{x}{2} \right )+^3C_2(1)\left ( \frac{x}{2} \right )^2+^3C_3\left ( \frac{x}{2} \right )^3

\left ( 1+\frac{x}{2} \right )^3= 1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^3}{8}..........(3)

Now, From (1), (2) and (3) we get,

\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4=1+2x+\frac{3x^2}{2}+\frac{x^3}{8}+\frac{x^4}{16}-\frac{8}{x}\left ( 1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^2}{8} \right )+\frac{8}{x^2}+\frac{24}{x}+6-\frac{32}{x^3}+\frac{16}{x^4}

\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4=1+2x+\frac{3x^2}{2}+\frac{x^3}{8}+\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}

\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4=1+2x+\frac{3x^2}{2}+\frac{x^3}{8}+\frac{x^4}{16}-\frac{8}{x}\left ( 1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^2}{8} \right )=\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

Posted by

Pankaj Sanodiya

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