Expand using Binomial Theorem

$$(1+ \dfrac{x}{2}-\dfrac{2}{x})^4,x\neq 0$$

Asked by Pragya Singh | 1 year ago |  120

##### Solution :-

$$^nC_0(1+\dfrac{x}{2})^4-\; ^nC_2(1+\dfrac{x}{2})^2-\;$$

$$^nC_3(1+\dfrac{x}{2})(\dfrac{2}{x})^3+\;^nC_4(\dfrac{2}{x})^4$$

$$(1+ \dfrac{x}{2})^4-4(1+ \dfrac{x}{2})^3(\dfrac{2}{x})$$

$$+6(1+x+ \dfrac{x^2}{4}) (\dfrac{4}{x^2})-4(1+ \dfrac{x}{2})$$

$$(\dfrac{8}{x^3})+\dfrac{16}{x^4}$$

$$(1+ \dfrac{x}{2})^4-\dfrac{8}{x}(1+ \dfrac{x}{2})^3$$

$$+\dfrac{24}{x^2}+\dfrac{24}{x}+6-\dfrac{32}{x^3}-\dfrac{16}{x^2}+\dfrac{16}{x^4}$$

$$(1+ \dfrac{x}{2})^4-\dfrac{8}{x}(1+ \dfrac{x}{2})^3$$

$$+\dfrac{8}{x^2}+\dfrac{24}{x}+6-\dfrac{32}{x^3}+\dfrac{16}{x^4}$$.............(1)

Again by using Binomial Theorem, we obtain

$$(1+ \dfrac{x}{2})^4=\;^4C_0(1)^4+\;^4C_1(1)^3\dfrac{x}{2}$$

$$+\;^4C_2(1)^2(\dfrac{x^2}{2})+\;^4C_3(1)^3(\dfrac{x}{2})^3+\;^4C_4(\dfrac{x}{2})^4$$

$$1+4\times \dfrac{x}{2}+6\times \dfrac{x^4}{4}\times +4\times \dfrac{x^3}{8}+\dfrac{x^3}{16}$$

$$1+2x+6 \dfrac{3x^2}{2}+\dfrac{x^3}{2}+\dfrac{x^4}{16}$$..............(2)

$$(1+ \dfrac{x}{2})^3=\;^3C_0(1)^3+\;^3C_2(1)^2(\dfrac{x}{2})$$

$$+\;^3C_3(1)\dfrac{x}{2}^3$$

$$1+\dfrac{3x}{2}+\dfrac{3x^2}{4}+\dfrac{x^3}{8}+\dfrac{x^3}{8}$$ ..........(3)

From (1), (2) and (3), we obtain

$$[ (1+\dfrac{x}{2})-\dfrac{2}{x}]^4$$

$$\dfrac{16}{x}+\dfrac{8}{x^2}-\dfrac{32}{x^3}+\dfrac{16}{x^4}-4x$$

$$+\dfrac{x^2}{2}+\dfrac{x^3}{2}+\dfrac{x^4}{16}-5$$

Answered by Pragya Singh | 1 year ago

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