Question

Expand the expression $(1-2x{)}^5$ ?

Answer 1

$\left({a+b}^n\right){=}_0^{n}c{a}^n{b}^0{+}_1^{n}c{a}^{n-1}{b}^1{+}_2^{n}c{a}^{n-2}{b}^2+....+...{.}_{n-1}^{n}c{a}^1{b}^{n-1}{+}_n^{n}c{a}^0{b}^n$

$\left({a+b}^5\right){=}_0^{5}c{a}^5{b}^0{+}_1^{5}c{a}^4{b}^1{+}_2^{5}c{a}^3{b}^2{+}_3^{5}c{a}^2{b}^3{+}_4^{5}ca{b}^4{+}_0^{5}c{a}^0{b}^5$

$\frac{5}{0(5-0)}{a}^5\times 1+\frac{5}{1(5-1)}{a}^4{b}^1+\frac{5}{2(5-2)}{a}^3{b}^2+\frac{5}{3(5-3)}{a}^2{b}^3+\frac{5}{4(5-4)}{ab}^4+\frac{5}{5(5-5)}{b}^5\times 1$

$\frac{5}{0\times 5}{a}^5+\frac{5}{1\times 4}{a}^{4}b+\frac{5}{23}{a}^3{b}^2+\frac{5}{32}{a}^2{b}^3+\frac{5}{41}{ab}^4+\frac{5}{50}{b}^5$

$\frac{5}{5}{a}^5+\frac{5\times 4}{1}{a}^{4}b+\frac{5\times 4\times 3}{23}{a}^3{b}^2+\frac{5\times 4\times 3}{2\times 1\times 3}{a}^3{b}^2+\frac{5\times 4\times 3}{3\times 1\times 3}{a}^2{b}^3+\frac{5\times 4}{4}{ab}^4+\frac{5}{5}{b}^5$

$a^5+5a^{4}b+{10a}^3{b}^2+{10a}^2{b}^3+{5ab}^4+b^4$

$\left({a+b}^5\right)=a^5+5a^{4}b+{10a}^3{b}^2+{10a}^2{b}^3+{5ab}^4+b^4$

$puta=1,b=-2x$

$(1-2x^5)=1^5+51^4\{-2x\}+101^3{\{-2x\}}^2+{101}^2{\{-2x\}}^3+{51\{-2x\}}^4+{\{-2x\}}^4$

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

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