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Question Number 114728 by aurpeyz last updated on 20/Sep/20

  (3+4x)^(−5)   =3^(−5) ×(1+(4/3)x)^(−5)   ...  pls what is the explanation behind the fact  that i always have to factorize to get the form  (1+b)^(−s)  any time i am solving for binomial  with negatuve power?  can it be solved without first factorizing?

$$ \\ $$$$\left(\mathrm{3}+\mathrm{4}{x}\right)^{−\mathrm{5}} \\ $$$$=\mathrm{3}^{−\mathrm{5}} ×\left(\mathrm{1}+\frac{\mathrm{4}}{\mathrm{3}}{x}\right)^{−\mathrm{5}} \\ $$$$... \\ $$$${pls}\:{what}\:{is}\:{the}\:{explanation}\:{behind}\:{the}\:{fact} \\ $$$${that}\:{i}\:{always}\:{have}\:{to}\:{factorize}\:{to}\:{get}\:{the}\:{form} \\ $$$$\left(\mathrm{1}+{b}\right)^{−{s}} \:{any}\:{time}\:{i}\:{am}\:{solving}\:{for}\:{binomial} \\ $$$${with}\:{negatuve}\:{power}? \\ $$$${can}\:{it}\:{be}\:{solved}\:{without}\:{first}\:{factorizing}? \\ $$

Commented by mr W last updated on 20/Sep/20

i don′t understand why you have a  problem with this:  (a+b)^n =a^n (1+(b/a))^n   certainly you don′t need to factorize!    (3+4x)^(−5) =Σ_(k=0) ^∞ C_4 ^(k+4) (3)^(−5−k) (−4x)^k     3^(−5) (1+((4x)/3))^(−5) =3^(−5) Σ_(k=0) ^∞ C_4 ^(k+4) (1)^(−5−k) (−((4x)/3))^k   =3^(−5) Σ_(k=0) ^∞ C_4 ^(k+4) (−((4x)/3))^k   =Σ_(k=0) ^∞ C_4 ^(k+4) 3^(−5−k) (−4x)^k

$${i}\:{don}'{t}\:{understand}\:{why}\:{you}\:{have}\:{a} \\ $$$${problem}\:{with}\:{this}: \\ $$$$\left({a}+{b}\right)^{{n}} ={a}^{{n}} \left(\mathrm{1}+\frac{{b}}{{a}}\right)^{{n}} \\ $$$${certainly}\:{you}\:{don}'{t}\:{need}\:{to}\:{factorize}! \\ $$$$ \\ $$$$\left(\mathrm{3}+\mathrm{4}{x}\right)^{−\mathrm{5}} =\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} \left(\mathrm{3}\right)^{−\mathrm{5}−{k}} \left(−\mathrm{4}{x}\right)^{{k}} \\ $$$$ \\ $$$$\mathrm{3}^{−\mathrm{5}} \left(\mathrm{1}+\frac{\mathrm{4}{x}}{\mathrm{3}}\right)^{−\mathrm{5}} =\mathrm{3}^{−\mathrm{5}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} \left(\mathrm{1}\right)^{−\mathrm{5}−{k}} \left(−\frac{\mathrm{4}{x}}{\mathrm{3}}\right)^{{k}} \\ $$$$=\mathrm{3}^{−\mathrm{5}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} \left(−\frac{\mathrm{4}{x}}{\mathrm{3}}\right)^{{k}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} \mathrm{3}^{−\mathrm{5}−{k}} \left(−\mathrm{4}{x}\right)^{{k}} \\ $$

Commented by aurpeyz last updated on 20/Sep/20

thanks Sir

$${thanks}\:{Sir} \\ $$

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