Question

Using binomial theorem, the expansion of ${\left(1+\frac{x}{2}-\frac{2}{x}\right)}^4,x\ne 0$, is

• A. $\frac{x^4}{16}+\frac{x^3}{2}+\frac{x^2}{2}-4x+\frac{16}{x}+\frac{8}{x^2}+\frac{32}{x^3}-\frac{16}{x^4}$
• B. $\frac{x^4}{16}+\frac{x^3}{2}+\frac{x^2}{2}-4x-5+\frac{16}{x}+8x^2+32x^3-16x^4$
• C. $\frac{x^4}{16}+\frac{x^3}{2}+\frac{x^2}{2}-4x-5+\frac{16}{x}+\frac{8}{x^2}-\frac{32}{x^3}+\frac{16}{x^4}$
• D. None of the above

Answer 1

Answer: (C)

${\left[\left(1+\frac{x}{2}\right)-\frac{2}{x}\right]}^4={}^4{C}_0{\left(1+\frac{x}{2}\right)}^4-{}^4{C}_1{\left(1+\frac{x}{2}\right)}^3\left(\frac{2}{x}\right)$
$+{}^4{C}_2{\left(1+\frac{x}{2}\right)}^2{\left(\frac{2}{x}\right)}^2-{}^4{C}_3\left(1+\frac{x}{2}\right){\left(\frac{2}{x}\right)}^3+{}^4{C}_4{\left(\frac{2}{x}\right)}^4$
$={\left(1+\frac{x}{2}\right)}^4-4{\left(1+\frac{x}{2}\right)}^3\frac{2}{x}+\frac{4\times 3}{2}{\left(1+\frac{x}{2}\right)}^2\frac{4}{x^2}$
$-4\left(1+\frac{x}{2}\right)\times \frac{8}{x^3}+\frac{16}{x^4}$
$={\left(1+\frac{x}{2}\right)}^4-\frac{8}{x}{\left(1+\frac{x}{2}\right)}^3+\frac{24}{x^2}{\left(1+\frac{x}{2}\right)}^2-\frac{32}{x^3}\left(1+\frac{x}{2}\right)+\frac{16}{x^4}$
Now, open the expansion of ${\left(1+\frac{x}{2}\right)}^4,{\left(1+\frac{x}{2}\right)}^3$ and
${\left(1+\frac{x}{2}\right)}^2$, we get
${\left(1+\frac{x}{2}-\frac{2}{x}\right)}^4=\left(1+4\cdot \frac{x}{2}+6\cdot \frac{x^2}{4}+4\cdot \frac{x^3}{8}+\frac{x^4}{16}\right)$
$-8\cdot \frac{1}{x}\left(1+3\cdot \frac{x}{2}+3\cdot \frac{x^2}{4}+\frac{x^3}{8}\right)$
$+24\cdot \frac{1}{x^2}\left(1+x+\frac{x^2}{4}\right)-32\times \frac{1}{x^3}\left(1+\frac{x}{2}\right)+\frac{16}{x^4}$
$=\left(1+2x+\frac{3x^2}{2}+\frac{x^3}{2}+\frac{x^4}{16}\right)-\left(\frac{8}{x}+12+6x+x^2\right)$
$+\left(\frac{24}{x^2}+\frac{24}{x}+6\right)-\left(\frac{32}{x^3}+\frac{16}{x^2}\right)+\frac{16}{x^4}$
$=\frac{x^4}{16}+\frac{x^3}{2}+x^2\left(\frac{3}{2}-1\right)+x(2-6)+(1-12+6)$
$+(24-8)\frac{1}{x}+(24-16)\frac{1}{x^2}-\frac{32}{x^3}+\frac{16}{x^4}$
$=\frac{x^4}{16}+\frac{x^3}{2}+\frac{x^2}{2}-4x-5+\frac{16}{x}+\frac{8}{x^2}-\frac{32}{x^3}+\frac{16}{x^4}$