Question

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

Answer 1

$\Rightarrow {(1+\frac{x}{2}-\frac{2}{x})}^4$

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

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

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

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

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

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

This is the final binomial expansion of above expression

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

$\Rightarrow 10000t=625$