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# Expand using Binomial theorem $(1+\large\frac{x}{2}-\frac{2}{x})^4$$\;\;x\neq 0 Can you answer this question? ## 1 Answer 0 votes Toolbox: • (a-b)^n=nC_0-nC_1a^{n-1}b+nC_2a^{n-2}b^2-......+(-1)^nnC_ra^{n-r}b^r+......+nC_n(-b)^n ((1+\large\frac{x}{2})-\frac{2}{x})^4=$$(1+\large\frac{x}{2})^3$$+4C_1(1+\large\frac{x}{2})^3(\large\frac{-2}{x})^2+$$4C_2(1+\large\frac{x}{2})^2(\large\frac{-2}{x})^2$$+4C_3(1+\large\frac{x}{2})(\large\frac{-2}{x})^3+$$4C_4(-\large\frac{2}{x})^4$
$\Rightarrow(1+\large\frac{x}{2})^4$$-8.\large\frac{1}{x}$$(1+\large\frac{x}{2})^3$$+24.\large\frac{1}{x^2}$$(1+\large\frac{x}{2})^2-$$32.\large\frac{1}{x^3}$$(1+\large\frac{x}{2})+\frac{16}{x^4}$
Expanding $(1+\large\frac{x}{2})^4,$$(1+\large\frac{x}{2})^3,$$(1+\large\frac{x}{2})^2$ we have $(1+\large\frac{x}{2}-\frac{2}{x})^4$
$(1+4.\large\frac{x}{2}$$+6.\large\frac{x^2}{4}$$+4.\large\frac{x^3}{8}+\frac{x^4}{16})-$$8.\large\frac{1}{x}$$(1+3.\large\frac{x}{2}$$+3.\large\frac{x^2}{4}$$+\large\frac{x^3}{8})+$$24.\large\frac{1}{x^2}$$(1+x+\large\frac{x^2}{4})$$-32\times \large\frac{1}{x^3}$$(1+\large\frac{x}{2})+\frac{16}{x^4}$
$\Rightarrow (1+2x+\large\frac{3}{2}$$x^2+\large\frac{1}{2}$$x^3+\large\frac{x^4}{16})-(\frac{8}{x}$$+12+6x+x^2)+(\large\frac{24}{x^2}+\frac{24}{x}+$$6)-(\large\frac{32}{x^3}+\frac{16}{x^2})+\frac{16}{x^4}$
$\Rightarrow \large\frac{x^4}{16}+\frac{x^3}{2}+\frac{x^2}{2}$$-4x-5+\large\frac{16}{x}+\frac{8}{x^2}-\frac{32}{x^3}+\frac{16}{x^4}$