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  4. Using binomial theorem, the expansion of (1+(x/2)-(2/x))4, x ≠ 0, is

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

Binomial Theorem

Solution:

$\left[\left(1+\frac{x}{2}\right)-\frac{2}{x}\right]^4={ }^4 C_0\left(1+\frac{x}{2}\right)^4-{ }^4 C_1\left(1+\frac{x}{2}\right)^3\left(\frac{2}{x}\right)$
$+{ }^4 C_2\left(1+\frac{x}{2}\right)^2\left(\frac{2}{x}\right)^2-{ }^4 C_3\left(1+\frac{x}{2}\right)\left(\frac{2}{x}\right)^3+{ }^4 C_4\left(\frac{2}{x}\right)^4$
$=\left(1+\frac{x}{2}\right)^4-4\left(1+\frac{x}{2}\right)^3 \frac{2}{x}+\frac{4 \times 3}{2}\left(1+\frac{x}{2}\right)^2 \frac{4}{x^2}$
$-4\left(1+\frac{x}{2}\right) \times \frac{8}{x^3}+\frac{16}{x^4}$
$=\left(1+\frac{x}{2}\right)^4-\frac{8}{x}\left(1+\frac{x}{2}\right)^3+\frac{24}{x^2}\left(1+\frac{x}{2}\right)^2-\frac{32}{x^3}\left(1+\frac{x}{2}\right)+\frac{16}{x^4}$
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)$
$+24 \cdot \frac{1}{x^2}\left(1+x+\frac{x^2}{4}\right)-32 \times \frac{1}{x^3}\left(1+\frac{x}{2}\right)+\frac{16}{x^4}$
$=\left(1+2 x+\frac{3 x^2}{2}+\frac{x^3}{2}+\frac{x^4}{16}\right)-\left(\frac{8}{x}+12+6 x+x^2\right)$
$+\left(\frac{24}{x^2}+\frac{24}{x}+6\right)-\left(\frac{32}{x^3}+\frac{16}{x^2}\right)+\frac{16}{x^4}$
$=\frac{x^4}{16}+\frac{x^3}{2}+x^2\left(\frac{3}{2}-1\right)+x(2-6)+(1-12+6)$
$+(24-8) \frac{1}{x}+(24-16) \frac{1}{x^2}-\frac{32}{x^3}+\frac{16}{x^4}$
$=\frac{x^4}{16}+\frac{x^3}{2}+\frac{x^2}{2}-4 x-5+\frac{16}{x}+\frac{8}{x^2}-\frac{32}{x^3}+\frac{16}{x^4}$