Question Number 114728 by aurpeyz last updated on 20/Sep/20
$$ \\ $$$$\left(\mathrm{3}+\mathrm{4}{x}\right)^{−\mathrm{5}} \\ $$$$=\mathrm{3}^{−\mathrm{5}} ×\left(\mathrm{1}+\frac{\mathrm{4}}{\mathrm{3}}{x}\right)^{−\mathrm{5}} \\ $$$$… \\ $$$${pls}\:{what}\:{is}\:{the}\:{explanation}\:{behind}\:{the}\:{fact} \\ $$$${that}\:{i}\:{always}\:{have}\:{to}\:{factorize}\:{to}\:{get}\:{the}\:{form} \\ $$$$\left(\mathrm{1}+{b}\right)^{−{s}} \:{any}\:{time}\:{i}\:{am}\:{solving}\:{for}\:{binomial} \\ $$$${with}\:{negatuve}\:{power}? \\ $$$${can}\:{it}\:{be}\:{solved}\:{without}\:{first}\:{factorizing}? \\ $$
Commented by mr W last updated on 20/Sep/20
$${i}\:{don}'{t}\:{understand}\:{why}\:{you}\:{have}\:{a} \\ $$$${problem}\:{with}\:{this}: \\ $$$$\left({a}+{b}\right)^{{n}} ={a}^{{n}} \left(\mathrm{1}+\frac{{b}}{{a}}\right)^{{n}} \\ $$$${certainly}\:{you}\:{don}'{t}\:{need}\:{to}\:{factorize}! \\ $$$$ \\ $$$$\left(\mathrm{3}+\mathrm{4}{x}\right)^{−\mathrm{5}} =\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} \left(\mathrm{3}\right)^{−\mathrm{5}−{k}} \left(−\mathrm{4}{x}\right)^{{k}} \\ $$$$ \\ $$$$\mathrm{3}^{−\mathrm{5}} \left(\mathrm{1}+\frac{\mathrm{4}{x}}{\mathrm{3}}\right)^{−\mathrm{5}} =\mathrm{3}^{−\mathrm{5}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} \left(\mathrm{1}\right)^{−\mathrm{5}−{k}} \left(−\frac{\mathrm{4}{x}}{\mathrm{3}}\right)^{{k}} \\ $$$$=\mathrm{3}^{−\mathrm{5}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} \left(−\frac{\mathrm{4}{x}}{\mathrm{3}}\right)^{{k}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} \mathrm{3}^{−\mathrm{5}−{k}} \left(−\mathrm{4}{x}\right)^{{k}} \\ $$
Commented by aurpeyz last updated on 20/Sep/20
$${thanks}\:{Sir} \\ $$