Using binomial theorem, expand each of the following:

To find: Expansion of $(1-2 x)^{5}$

Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$

(ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$

We have, $(1-2 x)^{5}$

$\Rightarrow\left[{ }^{5} \mathrm{C}_{0}(1)^{5}\right]+\left[{ }^{5} \mathrm{C}_{1}(1)^{5-1}(-2 \mathrm{x})^{1}\right]+\left[{ }^{5} \mathrm{C}_{2}(1)^{5-2}(-2 \mathrm{x})^{2}\right]+\left[{ }^{5} \mathrm{C}_{3}(1)^{5-3}(-2 \mathrm{x})^{3}\right]+\left[{ }^{5} \mathrm{C}_{4}(1)^{5-4}(-2 \mathrm{x})^{4}\right]+$

$\Rightarrow\left[\frac{5 !}{0 !(5-0) !}(1)^{5}\right]-\left[\frac{5 !}{1 !(5-1) !}(1)^{4}(2 x)\right]+\left[\frac{5 !}{2 !(5-2) !}(1)^{3}\left(4 x^{2}\right)\right]$

$-\left[\frac{5 !}{3 !(5-3) !}(1)^{2}\left(8 x^{3}\right)\right]+\left[\frac{5 !}{4 !(5-4) !}(1)^{1}\left(16 x^{4}\right)\right]-\left[\frac{5 !}{5 !(5-5) !}\left(32 x^{5}\right)\right]$

$\Rightarrow 1-5(2 x)+10\left(4 x^{2}\right)-10\left(8 x^{3}\right)+5\left(16 x^{4}\right)-1\left(32 x^{5}\right)$

$\Rightarrow 1-10 x+40 x^{2}-80 x^{3}+80 x^{4}-32 x^{5}$

On rearranging

Ans) $-32 x^{5}+80 x^{4}-80 x^{3}+40 x^{2}-10 x+1$