1 Answer
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
\([ (1+\dfrac{x}{2})-\dfrac{2}{x}]^4\)
= \( \dfrac{16}{x}+\dfrac{8}{x^2}-\dfrac{32}{x^3}+\dfrac{16}{x^4}-4x\)
\( +\dfrac{x^2}{2}+\dfrac{x^3}{2}+\dfrac{x^4}{16}-5\)
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