# Expand using Binomial Theorem.

Question:

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

Solution:

Using Binomial Theorem, the given expression $\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}$ can be expanded as

$\left[\left(1+\frac{x}{2}\right)-\frac{2}{x}\right]^{4}$

$={ }^{4} \mathrm{C}_{0}\left(1+\frac{\mathrm{x}}{2}\right)^{4}-{ }^{4} \mathrm{C}_{1}\left(1+\frac{\mathrm{x}}{2}\right)^{3}\left(\frac{2}{\mathrm{x}}\right)+{ }^{4} \mathrm{C}_{2}\left(1+\frac{\mathrm{x}}{2}\right)^{2}\left(\frac{2}{\mathrm{x}}\right)^{2}-{ }^{4} \mathrm{C}_{3}\left(1+\frac{\mathrm{x}}{2}\right)\left(\frac{2}{\mathrm{x}}\right)^{3}+{ }^{4} \mathrm{C}_{4}\left(\frac{2}{\mathrm{x}}\right)^{4}$

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

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

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

Again by using Binomial Theorem, we obtain

$\left(1+\frac{x}{2}\right)^{4}={ }^{4} C_{0}(1)^{4}+{ }^{4} C_{1}(1)^{3}\left(\frac{x}{2}\right)+{ }^{4} C_{2}(1)^{2}\left(\frac{x}{2}\right)^{2}+{ }^{4} C_{3}(1)^{1}\left(\frac{x}{2}\right)^{3}+{ }^{4} C_{4}\left(\frac{x}{2}\right)^{4}$

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

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

$\left(1+\frac{x}{2}\right)^{3}={ }^{3} C_{0}(1)^{3}+{ }^{3} C_{1}(1)^{2}\left(\frac{x}{2}\right)+{ }^{3} C_{2}(1)\left(\frac{x}{2}\right)^{2}+{ }^{3} C_{3}\left(\frac{x}{2}\right)^{3}$

$=1+\frac{3 x}{2}+\frac{3 x^{2}}{4}+\frac{x^{3}}{8}$ (3)

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

$\left[\left(1+\frac{x}{2}\right)-\frac{2}{x}\right]^{4}$

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

$=1+2 x+\frac{3}{2} x^{2}+\frac{x^{3}}{2}+\frac{x^{4}}{16}-\frac{8}{x}-12-6 x-x^{2}+\frac{8}{x^{2}}+\frac{24}{x}+6-\frac{32}{x^{3}}+\frac{16}{x^{4}}$

$=\frac{16}{x}+\frac{8}{x^{2}}-\frac{32}{x^{3}}+\frac{16}{x^{4}}-4 x+\frac{x^{2}}{2}+\frac{x^{3}}{2}+\frac{x^{4}}{16}-5$