Expand using Binomial Theorem ${\left(1+\frac{x}{2}-\frac{2}{x}\right)}^4,x\ne 0$ and let the sum of coefficients of the terms in the expansion be $t$. Find $10000t$ 

Let $$a=1+\dfrac {x} {2},\,b=\dfrac {2}{x}$$

$$\displaystyle \Rightarrow \binom {4}{0}\left (1+\dfrac {x} {2}\right) ^{4}-\binom {4}{1}\left (1+\dfrac {x} {2}\right) ^{3}\left (\dfrac {2}{x}\right) +\binom {4}{2}\left (1+\dfrac {x} {2}\right) ^{2}\left (\dfrac {2}{x}\right) ^{2}-\binom {4}{3}\left (1+\dfrac {x} {2}\right) ^{1}\left (\dfrac {2}{x}\right) ^{3}+\binom {4}{4}\left (\dfrac {2}{x}\right) ^{4}$$

$$\Rightarrow \left (1+\dfrac {x} {2}\right) ^{4}-\dfrac {8}{x}\left (1+\dfrac {x} {2}\right) ^{3} +\dfrac {24}{x^{2}}\left (1+\dfrac {x} {2}\right) ^{2}-\dfrac {32}{x^{3}}\left (1+\dfrac {x} {2}\right) ^{1}+\left (\dfrac {2}{x}\right) ^{4}$$

We will use binomial expansion for $$\left (1+\dfrac {x} {2}\right) ^{4},\,\left (1+\dfrac {x} {2}\right) ^{3}$$

$$\Rightarrow \left [1+4. \dfrac {x} {2}+6. \dfrac {x^{2}}{4}+4. \dfrac {x^{3}}{8}+\dfrac {x^{4}}{16}\right] - \dfrac {8}{x}\left [1+3.\dfrac {x} {2}+3\dfrac {x^{2}}{4}+\dfrac {x^{3}}{8}\right]+\dfrac {24}{x^{2}}\left [1+\dfrac {x^{2}}{4}+x\right] -\dfrac{32}{x^{3}}\left[1+\dfrac{x}{2}\right]+\dfrac {16}{x^{4}}$$

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

This is the final binomial expansion of above expression 

$$\therefore$$sum of coefficients=$$\dfrac{1}{16}+\dfrac{1}{2}+\dfrac{1}{2}-4-5+16+8-32+16=\dfrac{1}{16}=t$$

$$\Rightarrow10000t=625$$